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Applications of Derivatives IITJEE Maths tutorial: practice problems part IV

November 13, 2018 – 6:26 am

Question 1.

If the point on $y = x \tan {\alpha} - \frac{ax^{2}}{2u^{2}\cos^{2}{\alpha}}$ , where $\alpha>0$ , where the tangent is parallel to $y=x$ has an ordinate $\frac{u^{2}}{4a}$ , then what is the value of $\alpha$ ?

Question 2:

Prove that the segment of the tangent to the curve $y=c/x$ , which is contained between the coordinate axes is bisected at the point of tangency.

Question 3:

Find all the tangents to the curve $y = \cos{(x+y)}$ for $-\pi \leq x \leq \pi$ that are parallel to the line $x+2y=0$ .

Question 4:

Prove that the curves $y=f(x)$ , where $f(x)>0$ , and $y=f(x)\sin{x}$ , where $f(x)$ is a differentiable function have common tangents at common points.

Question 5:

Find the condition that the lines $x \cos{\alpha} + y \sin{\alpha} = p$ may touch the curve $(\frac{x}{a})^{m} + (\frac{y}{b})^{m}=1$ .

Question 6:

Find the equation of a straight line which is tangent to one point and normal to the point on the curve $y=8t^{3}-1$ , and $x=4t^{2}+3$ .

Question 7:

Three normals are drawn from the point $(c,0)$ to the curve $y^{2}=x$ . Show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the two other normals are perpendicular to each other.

Question 8:

If $p_{1}$ and $p_{2}$ are lengths of the perpendiculars from origin on the tangent and normal to the curve $x^{2/3} + y^{2/3}=a^{2/3}$ respectively, prove that $4p_{1}^{2} + p_{2}^{2}=a^{2}$ .

Question 9:

Show that the curve $x=1-3t^{2}$ , and $y=t-3t^{3}$ is symmetrical about x-axis and has no real points for $x>1$ . If the tangent at the point t is inclined at an angle $\psi$ to OX, prove that $3t= \tan {\psi} +\sec {\psi}$ . If the tangent at $P(-2,2)$ meets the curve again at Q, prove that the tangents at P and Q are at right angles.

Question 10:

Find the condition that the curves $ax^{2}+by^{2}=1$ and $a^{'}x^{2} + b^{'}y^{2}=1$ intersect orthogonality and hence show that the curves $\frac{x^{2}}{(a^{2}+b_{1})} + \frac{y^{2}}{(b^{2}+b_{1})} = 1$ and $\frac{x^{2}}{a^{2}+b_{2}} + \frac{y^{2}}{(b^{2}+b_{2})} =1$ also intersect orthogonally.

More later,

Nalin Pithwa.

By Nalin Pithwa | Also posted in applications of maths, calculus, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Applications of Derivatives: Tutorial: IITJEE Maths: Part II

October 23, 2018 – 8:32 pm

Another set of “easy to moderately difficult” questions:

The function $y = \frac{}x{1+x^{2}}$ decreases in the interval (a) $(-1,1)$ (b) $[1, \infty)$ (c) $(-\infty, -1]$ (d) $(-\infty, \infty)$ . There are more than one correct choices. Which are those?
The function $f(x) = \arctan (x) - x$ decreases in the interval (a) $(1,\infty)$ (b) $(-1, \infty)$ (c) $(-\infty, -\infty)$ (d) $(0, \infty)$ . There is more than one correct choice. Which are those?
For $x>1$ , $y = \log(x)$ satisfies the inequality: (a) $x-1>y$ (b) $x^{2}-1>y$ (c) $y>x-1$ (d) $\frac{x-1}{x}<y$ . There is more than one correct choice. Which are those?
Suppose $f^{'}(x)$ exists for each x and $h(x) = f(x) - (f(x))^{2} + (f(x))^{3}$ for every real number x. Then, (a) h is increasing whenever f is increasing (b) h is increasing whenever f is decreasing (c) h is decreasing whenever f is decreasing (d) nothing can be said in general. Find the correct choice(s).
If $f(x)=3x^{2}+12x-1$ , when $-1 \leq x \leq 2$ , and $f(x)=37-x$ , when $2<x\leq 3$ . Then, (a) $f(x)$ is increasing on $[-1,2]$ (b) $f(x)$ is continuous on $[-1,3]$ (c) $f^{'}(2)$ doesn’t exist (d) $f(x)$ has the maximum value at $x=2$ . Find all the correct choice(s).
In which interval does the function $y=\frac{x}{\log(x)}$ increase?
Which is the larger of the functions $\sin(x) + \tan(x)$ and $f(x)=2x$ in the interval $(0<x<\pi/2)$ ?
Find the set of all x for which $\log {(1+x)} \leq x$ .
Let $f(x) = |x-1| + a$ , if $x \leq 1$ ; and, $f(x)=2x+3$ , if $x>1$ . If $f(x)$ has local minimum at $x=1$ , then $a \leq$ ?
There are exactly two distinct linear functions (find them), such that they map $[-1,1]$ and $[0,2]$ .

more later, cheers,

Nalin Pithwa.

By Nalin Pithwa | Also posted in applications of maths, calculus | Comments (0)

Applications of Derivatives: A Quick Review

July 17, 2018 – 12:19 am

Section I:

The Derivative as a Rate of Change

In case of a linear function $y=mx+c$ , the graph is a straight line and the slope m measures the steepness of the line by giving the rate of climb of the line, the rate of change of y with respect to x.

As x changes from $x_{0}$ to $x_{1}$ , y changes m times as much:

$y_{1}-y_{0}=m(x_{1}-x_{0})$

Thus, the slope $m=(y_{1}-y_{0})(x_{1}-x_{0})$ gives the change in y per unit change in x.

In more general case of differentiable function $y=f(x)$ , the difference quotient

$\frac{f(x+h)-f(x)}{x+h-x} = \frac{f(x+h)-f(x)}{h}$ , where $h \neq 0$

give the average rate of change of y (or f) with respect to x. The limit as h approaches zero is the derivative $dy/dx = f^{'}(x)$ , which can be interpreted as the instantaneous rate of change of f with respect to x. Since, the graph is a curve, the rate of change of y can vary from point to point.

Velocity and Acceleration:

Suppose that an object is moving along a straight line and that, for each time t during a certain time interval, the object has location/position $x(t)$ . Then, at time $t+h$ the position of the object is $x(t+h)$ and $x(t+h)-x(t)$ is the change in position that the object experienced during the time period t to $t+h$ . The ratio

$\frac{x(t+h)-x(t)}{t+h-t} = \frac{x(t+h)-x(t)}{h}$

gives the average velocity of the object during this time period. If

$\lim_{h \rightarrow 0} \frac{x(t+h)-x(t)}{h}=x^{'}(t)$

exists, then $x^{'}(t)$ gives the instantaneous rate of change of position with respect to time. This rate of change of position is called the velocity of the object. If the velocity function is itself differentiable, then its rate of change with respect to time is called the acceleration; in symbols,

$a(t) = v^{'}(t) = x^{''}(t)$

The speed is by definition the absolute value of the velocity: speed at time t is $|v(t)|$

If the velocity and acceleration have the same sign, then the object is speeding up, but if the velocity and acceleration have opposite signs, then the object is slowing down.

A sudden change in acceleration is called a jerk. Jerk is the derivative of acceleration. If a body’s position at the time t is $x(t)$ , the body’s jerk at time t is

$j = \frac{da}{dt} = \frac{d^{3}x}{dt^{3}}$

Differentials

Let $y = f(x)$ be a differentiable function. Let $h \neq 0$ . The difference $f(x+h) - f(x)$ is called the increment of f from x to $x+h$ , and is denoted by $\Delta f$ .

$\Delta f = f(x+h) - f(x)$

The product $f^{'}(x)h$ is called the differential of f at x with increment h, and is denoted by $df$

$df = f^{'}(x)h$

The change in f from x to $x+h$ can be approximated by $f^{'}(x)h$ :

$f(x+h) - f(x) = f^{'}(x)h$

Tangent and Normal

Let $y = f(x)$ be the equation of a curve, and let $P(x_{0}, y_{0})$ be a point on it. Let PT be the tangent, PN the normal and PM the perpendicular to the x-axis.

The slope of the tangent to the curve $y = f(x)$ at P is given by $(\frac{dy}{dx})_{(x_{0}, y_{0})}$

Thus, the equation of the tangent to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is $y - y_{0} = (\frac{dy}{dx})_{(x_{0}, y_{0})}(x-x_{0})$

Since PM is perpendicular to PT, it follows that if $(\frac{dy}{dx})_{(x_{0}, y_{0})} \neq 0$ , the slope of PN is

$- \frac{1}{(\frac{dy}{dx})_{(x_{0}, y_{0})}} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}$

Hence, the equation of the normal to the curve $y = f(x)$ at $(x_{0}, y_{0})$ is

$y - y_{0} = - (\frac{dx}{dy})_{(x_{0}, y_{0})}(x-x_{0})$

The equation of the normal parallel to the x-axis is $y = y_{0}$ , that is, when $(\frac{dy}{dx})_{(x_{0}, y_{0})} = 0$ . The length of the tangent at $(x_{0}, y_{0})$ is PT, and it is equal to

$y_{0}\csc{\theta} = y_{0}\sqrt{1+\cot^{2}{\theta}} = y_{0}\sqrt{1+[(\frac{dx}{dy})_{(x_{0}, y_{0})}]^{2}}$

The length of the normal is PN and it is equal to $y_{0}\sec {\theta} = y_{0}\sqrt{1 + [(\frac{dy}{dx})_{(x_{0}, y_{0})}]^{2}}$

If the curve is represented by $x = f(t)$ and $y = g(t)$ , that is, parametric equations in t, then

$\frac{dy}{dx} = \frac{g^{'}(t)}{f^{'}(t)}$ where $g^{'}(t)= \frac{dy}{dt}$ and $f^{'}(t) = \frac{dx}{dt}$ . In this case, the equations of the tangent and the normal are given by

$y - g(t) = \frac{g^{'}(t)}{f^{'}(t)}[x - f(t)]$ and $[y-g(t)] g^{'}(t) + [x-f(t)]f^{'}(t) = 0$ respectively.

The Angle between Two Curves

The angle of intersection of two curves is defined as the angle between the two tangents at the point of intersection. Let $y = f(x)$ and $y=g(x)$ be two curves, and let $P(x_{0}, y_{0})$ be their point of intersection. Also, let $\psi$ and $\phi$ be the angles of inclination of the two tangents with the x-axis, and let $\theta$ be the angle between the two tangents. Then,

$\tan {\theta} = \frac{\tan{\phi}-\tan{\psi}}{1+\tan{\phi}\tan{\psi}} = \frac{g^{'}(x) - f^{'}(x)}{1+f^{'}(x)g^{'}(x)}$

Example 1:

Write down the equations of the tangent and the normal to the curve $y = x^{3} - 3x + 2$ at the point $(2,4)$ .

Solution 1:

$\frac{dy}{dx} = 3x^{2}-3 \Longrightarrow \frac{dy}{dx}_{(2,4)} = 3.4 - 3 = 9$ .

Hence, the equation of the tangent at $(2,4)$ is given by $y-4 = 9(x-2) \Longrightarrow 9x-y-14=0$ and the equation of the normal is $y - 4 = (-1/9)(x-2) \Longrightarrow x+9y -38=0$ .

Rolle’s Theorem and Lagrange’s Theorem:

Rolle’s Theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) f(x) is continuous on $[a,b]$ , (ii) f(x) is derivable on $(a,b)$ , and (iii) f(a) = f(b). Then, there exists a $c \in (a,b)$ such that $f^{'}(x)=0$ .

For details, the very beautiful, lucid, accessible explanation in Wikipedia:

https://en.wikipedia.org/wiki/Rolle%27s_theorem

Lagrange’s theorem:

Let $f(x)$ be a function defined on a closed interval $[a,b]$ such that (i) $f(x)$ is continuous on $[a,b]$ , and (ii) $f(x)$ is derivable on $(a,b)$ . Then, there exists a $c \in [a,b]$ such that

$f^{'}(c) = \frac{f(b)-f(a)}{b-a}$

Example 2:

The function $f(x) = \log {\sin(x)}$ satisfies the conditions of Rolle’s theorem on the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$ , as the logarithmic function and $\sin (x)$ are continuous and differentiable functions and $\log {\sin (\frac{5\pi}{6})} = \log {\sin (\pi - \frac{\pi}{6})} = \log{\sin{(\frac{\pi}{6})}}$ .

The conclusion of Rolle’s theorem is given at $c=\frac{\pi}{2}$ , for which $f^{'}(c) = \cot (c) = \cot (\pi/2) =0$ .

Rolle’s theorem for polynomials:

If $\phi(x)$ is any polynomial, then between any pair of roots of $\phi(x)=0$ lies a root of $\phi^{'}(x)=0$ .

Monotonicity:

A function $f(x)$ defined on a set D is said to be non-decreasing, increasing, non-increasing and decreasing respectively, if for any $x_{1}, x_{2} \in D$ and $x_{1} < x_{2}$ , we have $f(x_{1}) \leq f(x_{2})$ , $f(x_{1}) < f(x_{2})$ , $f(x_{1}) \geq f(x_{2})$ and $f(x_{1}) > f(x_{2})$ respectively. The function $f(x)$ is said to be monotonic if it possesses any of these properties.

For example, $f(x) = e^{x}$ is an increasing function, and $f(x)=\frac{1}{x}$ is a decreasing function.

Testing monotonicity:

Let $f(x)$ be continuous on $[a,b]$ and differentiable on $(a,b)$ . Then,

(i) for $f(x)$ to be non-decreasing (non-increasing) on $[a,b]$ it is necessary and sufficient that $f^{'}(x) \geq 0$ ( $f^{'}(x) \leq 0$ ) for all $x \in (a,b)$ .

(ii) for $f(x)$ to be increasing (decreasing) on $[a,b]$ it is sufficient that $f^{'}(x)>0$ ( $f^{'}(x)<0$ ) for all $x \in (a,b)$ .

(iii) If $f^{'}(x)=0$ for all x in $(a,b)$ , then f is constant on $[a,b]$ .

Example 3:

Determine the intervals of increase and decrease for the function $f(x)=x^{3}+2x-5$ .

Solution 3:

We have $f^{'}(x) = 3x^{2}+2$ , and for any value of x, $3x^{2}+2>0$ . Hence, f is increasing on $(-\infty, -\infty)$ . QED.

The following is a simple criterion for determining the sign of $f^{'}(x)$ :

If $a,b \geq 0$ , then $(x-a)(x-b)>0$ iff $x > \max (a,b)$ or $x < \min(a,b)$ ;

$(x-a)(x-b)<0$ if and only if $\min(a,b) < x < \max(a,b)$

Maxima and Minima:

A function has a local maximum at the point $x_{0}$ if the value of the function $f(x)$ at that point is greater than its values at all points other than $x_{0}$ of a certain interval containing the point $x_{0}$ . In other words, a function $f(x)$ has a maximum at $x_{0}$ if it is possible to find an interval $(\alpha, \beta)$ containing $x_{0}$ , that is, with $\alpha < x_{0} < \beta$ , such that for all points different from $x_{0}$ in $(\alpha, \beta)$ , we have $f(x) < f(x_{0})$ .

A function $f(x)$ has a local minimum at $x_{0}$ if there exists an interval $(\alpha, \beta)$ containing $x_{0}$ such that $f(x) > f(x_{0})$ for $x \in (\alpha, \beta)$ and $x \neq x_{0}$ .

One should not confuse the local maximum and local minimum of a function with its largest and smallest values over a given interval. The local maximum of a function is the largest value only in comparison to the values it has at all points sufficiently close to the point of local maximum. Similarly, the local minimum is the smallest value only in comparison to the values of the function at all points sufficiently close to the local minimum point.

The general term for the maximum and minimum of a function is extremum, or the extreme values of the function. A necessary condition for the existence of an extremum at the point $x_{0}$ of the function $f(x)$ is that $f^{'}(x_{0})=0$ , or $f^{'}(x_{0})$ does not exist. The points at which $f^{'}(x)=0$ or $f^{'}(x)$ does not exist, are called critical points.

First Derivative Test:

(i) If $f^{'}(x)$ changes sign from positive to negative at $x_{0}$ , that is, $f^{'}(x)>0$ for $x < x_{0}$ and $f^{'}(x)<0$ for $x > x_{0}$ , then the function attains a local maximum at $x_{0}$ .

(ii) If $f^{'}(x)$ changes sign from negative to positive at $x_{0}$ , that is, $f^{'}(x)<0$ for $x<x_{0}$ , and $f^{'}(x)>0$ for $x > x_{0}$ , then the function attains a local minimum at $x_{0}$ .

(iii) If the derivative does not change sign in moving through the point $x_{0}$ , there is no extremum at that point.

Second Derivative Test:

Let f be twice differentiable, and let c be a root of the equation $f^{'}(x)=0$ . Then,

(i) c is a local maximum point if $f^{''}(c)<0$ .

(ii) c is a local minimum point if $f^{''}(c)>0$ .

However, if $f^{''}(c)=0$ , then the following result is applicable. Let $f^{'}(c) = f^{''}(c) = \ldots = f^{n-1}(c)=0$ (where f^{r} denotes the rth derivative), but $f^{(n)}(c) \neq 0$ .

(i) If n is even and $f^{(n)}(c)<0$ , there is a local maximum at c, while if $f^{(n)}(c)>0$ , there is a local minimum at c.

(ii) If n is odd, there is no extremum at the point c.

Greatest/Least Value (Absolute Maximum/Absolute Minimum):

Let f be a function with domain D. Then, f has a greatest value (or absolute maximum) at a point $c \in D$ if $f^(x) \leq f(c)$ for all x in D and a least value (or absolute minimum) at c, if $f(x) \geq f(c)$ for all x in D.

If f is continuous at every point of D, and $D=[a,b]$ , a closed interval, the f assumes both a greatest value M and a least value m, that is, there are $x_{1}, x_{2} \in [a,b]$ such that $f(x_{1})=M$ and $f(x_{2})=m$ , and $m \leq f(x) \leq M$ for every $x \in [a,b]$ .

Example 4:

a) $y=x^{2}$ , with domain $(-\infty, \infty)$ . This has no greatest value; least value at $x=0$

b) $y=x^{2}$ with domain $[0,2]$ . This has greatest value at $x=2$ and least value at $x=0$ .

c) $y=x^{2}$ with domain $(0,2]$ . This has greatest value at $x=2$ and no least value.

d) $y=x^{2}$ with domain $(0,2)$ . This has no greatest value and no least value.

Some other remarks:

The greatest (least) value of continuous function $f(x)$ on the interval $[a,b]$ is attained either at the critical points or at the end points of the interval. To find the greatest (least) value of the function, we have to compute its values at all the critical points on the interval $(a,b)$ , and the values $f(a), f(b)$ of the function at the end-points of the interval, and choose the greatest (least) out of the values so obtained.

We will continue with problems on applications of derivatives later,

Nalin Pithwa.

By Nalin Pithwa | Also posted in applications of maths, calculus | Comments (0)

Co-ordinate geometry practice for IITJEE Maths: Ellipses

May 6, 2018 – 11:19 pm

Problem 1:

Find the locus of the point of intersection of tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}}=1$ , which are at right angles.

Solution I:

Any tangent to the ellipse is $y = mx + \sqrt{a^{2}m^{2}+b^{2}}$ ….call this equation I.

Equation of the tangent perpendicular to this tangent is $y=-\frac{-1}{m}x+\sqrt{\frac{a^{2}}{m^{2}}+b^{2}}$ …call this equation II.

The locus of the intersection of tangent lines (i) and (ii) is obtained by eliminating m between these equations. So, squaring and adding them, we get

$(y-mx)^{2}+(my+x)^{2}=a^{2}m^{2}+b^{2}+a^{2}+b^{2}m^{2}$

$\Longrightarrow (1+m^{2})(x^{2}+y^{2})=(1+m^{2})(a^{2}+b^{2})$

$\Longrightarrow x^{2}+y^{2}=a^{2}+b^{2}$

which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the director circle of the ellipse.

Problem II:

A tangent to the ellipse $x^{2}+4y^{2}=4$ meets the ellipse $x^{2}+2y^{2}=6$ at P and Q. Prove that the tangents at P and Q of the ellipse $x^{2}+2y^{2}=6$ at right angles.

Solution II:

Let the tangent at $R(2\cos {\theta}, \sin{\theta})$ to the ellipse $x^{2}+4y^{2}=4$ meet the ellipse $x^{2}+2y^{2}=6$ at P and Q.

Let the tangents at P and Q to the second ellipse intersect at the point $S(\alpha,\beta)$ . Then, PQ is the chord of contact of the point $S(\alpha,\beta)$ with respect to ellipse two, and so its equation is

$\alpha x + 2\beta y=6$ ….call this “A”.

PQ is also the tangent at $R(2\cos {\theta}, \sin{\theta})$ to the first ellipse and so the equation can be written as $(2\cos{\theta})x+(4\sin{\theta})y=4$ ….call this “B”.

Comparing “A” and “B”, we get $\frac{2\cos{\theta}}{\alpha} = \frac{4\sin{\theta}}{2\beta} = \frac{4}{6}$

$\Longrightarrow \cos{\theta}=\frac{\alpha}{3}$ and $\sin{\theta}=\frac{\beta}{3}$

$\Longrightarrow \frac{\alpha^{2}}{9} + \frac{\beta^{2}}{9}=1 \Longrightarrow \alpha^{2}+\beta^{2}=9$

The locus of $S( \alpha, \beta)$ is $x^{2}+y^{2}=9$ , or $x^{2}+y^{2}=6+3$ , which is the director circle of the second ellipse.

Hence, the tangents at P and Q to the ellipse (ii) are at right angles (by the solution to the previous example).

Problem 3:

Let d be the perpendicular distance from the centre of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ to the tangent drawn at a point P on the ellipse. If $F_{1}$ and $F_{}{2}$ are the two foci of the ellipse, then show that $(PF_{1}-PF_{2})^{2}=4a^{2}(1-\frac{b^{2}}{d^{2}})$

Solution 3:

Equation of the tangent at the point $P(a\cos {\theta}, b\sin{\theta})$ on the given ellipse is $\frac{x\cos{\theta}}{a} + \frac{y\sin{\theta}}{b}=1$ . Thus,

$d= |\frac{-1}{\sqrt{\frac{\cos^{2}{\theta}}{a^{2}}+\frac{\sin^{2}{\theta}}{b^{2}}}}|$

$\Longrightarrow d^{2}=\frac{a^{2}b^{2}}{b^{2}\cos^{2}{\theta}+a^{2}\sin^{2}{\theta}}$

We know $PF_{1}+PF_{2}=2a$

$\Longrightarrow (PF_{1}-PF_{2})^{2}=(PF_{1}+PF_{2})^{2}-4PF_{1}PF_{2}$ …call this equation I.

Also, $(PF_{1}.PF_{2})^{2}=[ (a\cos{\theta}-ae)^{2}+(b\sin{\theta})^ {2}].[(a\cos{\theta}+ae)^{2} + (b\sin{\theta})^{2}]$ , which in turn equals,

$[a^{2}(\cos{\theta}-e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]. [a^{2}(\cos{\theta}+e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]$ , that is,

$a^{4}[ (\cos^{2}{\theta}+e^{2}) -2e\cos{\theta}+\sin^{2}{\theta} - e^{2} \sin^{2}{\theta} ]. [ (\cos^{2}{\theta}+e^{2}) + 2e \cos{\theta} + \sin^{2}{\theta} - e^{2}\sin^{2}{\theta}] = a^{4}[ 1-2e\cos{\theta}+e^{2}\cos^{2}{\theta} ] [1+ 2e \cos{\theta} + e^{2}\cos^{2}{\theta} ]$

that is,

$a^{4}[ (1+e^{2}\cos^{2}{\theta})^{2}-4e^{2}\cos^{2}{\theta}] = a^{4}(1-e^{2}\cos^{2}{\theta})^{2}$

$\Longrightarrow PF_{1}. PF_{2}=a^{2}(1-e^{2}\cos^{2}{\theta})$

Now, from I, we get $(PF_{1} - PF_{2})^{2} = 4a^{2}-4a^{2}(1-e^{2}\cos^{2}{\theta}) = 4a^{2}e^{2}\cos^{2}{\theta}$ ,

also, $1-\frac{b^{2}}{a^{2}} = 1 - \frac{b^{2}\cos^{2}{\theta} + a^{2}\sin^{2}{\theta}}{a^{2}} = \frac{(a^{2}-b^{2})\cos^{2}{\theta}}{a^{2}} = e^{2}\cos^{2}{\theta}$

Hence, $(PF_{1} - PF_{2})^{2} = 4a^{2}(1-\frac{b^{2}}{d^{2}})$

we will continue later,

Cheers,

Nalin Pithwa

By Nalin Pithwa | Also posted in co-ordinate geometry | Comments (0)

Co-ordinate Geometry problems for IITJEE : equations of median, area of a triangle, and circles

April 30, 2018 – 9:54 pm

Problem I:

If $A(x_{1}, y_{1})$ , $B(x_{1}, y_{1})$ and $C(x_{3}, y_{3})$ are the vertices of a triangle ABC, then prove that the equation of the median through A is given by:

$\left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right | + \left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right |=0$

Solution I:

If D is the mid-point of BC, its co-ordinates are $( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} )$

Therefore, equation of the median AD is $\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1\\ \frac{x_{2}+x_{3}}{2} & \frac{y_{2}+y_{3}}{2} & 1 \end{array} \right|=0$ , which in turn, implies that,

$\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2}+x_{3} & y_{2}+y_{3} & 2 \end{array}\right |=0$

Now apply the row transformation $R_{3} \rightarrow 2R_{3}$ to the previous determinant. So, we get

$\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right | + \left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right |=0$ , using the sum property of determinants.

Hence, the proof.

Problem 2:

If $\triangle_{1}$ is the area of the triangle with vertices $(0,0), (a\tan {\alpha},b\cot{\alpha}), (a\sin{\alpha}, b\cos {\alpha})$ , and $\triangle_{2}$ is the area of the triangle with vertices $(a,b)$ , $(a\sec^{2}{\alpha}, b\csc^{2}{\alpha})$ , and $(a+a\sin^{2}{\alpha}, b+b\cos^{2}{\alpha})$ , and $\triangle_{3}$ is the area of the triangle with vertices $( 0, 0)$ , $( a\tan{\alpha}, -b\cos{\alpha})$ , $(a\sin{\alpha},b\cos{\alpha})$ . Then, prove that there is no value of $\alpha$ for which the areas of triangles, $\triangle_{1}$ , $\triangle_{2}$ and $\triangle_{3}$ are in GP.

Solution 2:

We have $\triangle_{1}=\frac{1}{2}|\left | \begin{array}{ccc}0 & 0 & 1 \\ a\tan{\alpha} & b\cot {\alpha} & 1 \\ a \sin{\alpha} & b\cos{\alpha} & 1 \end{array}\right ||=\frac{1}{2}ab|\sin{\alpha}-\cos{\alpha}|$ , and

$\triangle_{2}=\frac{1}{2}|\left | \begin{array}{ccc}a & b & 1 \\ a\sec^{2}{\alpha} & b\csc^{2}{\alpha} & 1 \\ a + a\sin^{2}{\alpha} & b + b\cos^{2}{\alpha} & 1 \end{array} \right | |$ .

Applying the following column transformations to the above determinant, $C_{1} \rightarrow -aC_{3}$ and $C_{2}-bC_{3}$ , we get

$\triangle_{2}=\frac{1}{2}ab\left | \begin{array}{ccc}0 & 0 & 1 \\ \tan^{2}{\alpha} & \cot^{2}{\alpha} & 1 \\ \sin^{2}{\alpha} & \cos^{2}{\alpha} & 1 \end{array}\right | = \frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha})$ and $\triangle_{3}=\frac{1}{2}|\left | \begin{array}{ccc} 0 & 0 & 1 \\ a\tan{\alpha} & -b\cot{\alpha} & 1 \\ a\sin{\alpha} & b\cos{\alpha} & 1 \end{array} \right | |=\frac{1}{2}ab |\sin {\alpha}+\cos{\alpha}|$

so that $\triangle_{1}\triangle_{3}=\frac{1}{2}ab\triangle_{2}$ .

Now, $\triangle_{1}$ , $\triangle_{2}$ and $\triangle_{3}$ are in GP, if $\triangle_{1}\triangle_{3}=\triangle_{2}^{2} \Longrightarrow \frac{1}{2}ab\triangle_{2}=\triangle_{2}^{2} \Longrightarrow \triangle_{2}=\frac{1}{2}ab$

$\Longrightarrow \triangle_{2}=\frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha})=\frac{1}{2}ab \Longrightarrow (\sin^{2}{\alpha}-\cos^{2}{alpha})=1$ , that is,

$\alpha = (2m+1)\pi/2$ , where $m \in Z$ . But, for this value of $\alpha$ , the vertices of the given triangles are not defined. Hence, $\triangle_{1}$ , and $\triangle_{2}$ and $\triangle_{3}$ cannot be in GP for any value of $\alpha$ .

Problem 3:

Two points P and Q are taken on the line joining the points $A(0,0)$ and $B(3a,0)$ such that $AP=PQ=QB$ . Circles are drawn on AP, PQ, and QB as diameters. The locus of the point S, the sum of the squares of the length of the tangents from which to the three circles is equal to $b^{2}$ , is

(a) $x^{2}+y^{2}-3ax+2a^{2}-b^{2}=0$

(b) $3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0$

(d) $x^{2}+y^{2}-ax-b^{2}=0$ .

Ans. b.

Solution 3:

Since $AP=PQ=QB$ , the co-ordinates of P are $(a,0)$ and of Q are $(2a,0)$ , equations of the circles on AP, PQ, and QB as diameters are respectively.

Please draw the diagram.

So, we get

$(x-0)(x-a)+y^{2}=0$

$(x-a)(x-2a)+y^{2}=0$

$(x-2a)(x-3a)+y^{2}=0$

So, if $(h,k)$ be any point of the locus, then $3(h^{2}+k^{2})-9ah+8a^{2}=b^{2}$ .

So, the required locus of $(h,k)$ is $3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0$ .

More later,

Nalin Pithwa.

By Nalin Pithwa | Also posted in co-ordinate geometry | Tagged areas of triangles, co-ordinate geometry, GP, median | Comments (0)

Co-ordinate Geometry : IITJEE Mains practice: some random problems again

March 13, 2018 – 12:35 pm

Problem 1:

The line $Ax+By+C=0$ cuts the circle $x^{2}+y^{2}+ax+by+c=0$ at P and Q. The line $A^{'}x+B^{'}y+C^{'}=0$ cuts the circle $x^{2}+y^{2}+a^{'}x+b^{'}y+c^{'}=0$ at R and S. If P, Q, R and S are concyclic, prove that

$\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0$ .

Solution I;

An equation of a circle through P and Q is $x^{2}+y^{2}+ax+by+c +\lambda (Ax+By+C)=0$ …call this equation I.

And, an equation of a circle through R and S is $x^{2}+y^{2}+a^{'}x+b^{'}y + c^{'}+\mu (A^{'}x+B^{'}y+C^{'})=0$ …call this equation II.

If P, Q, R and S are concyclic, then I and II represent the same circle for same values of $\lambda$ and $\mu$ .

$\Longrightarrow a+ \lambda A=a^{'}+\mu A^{'}$ or $a-a^{'} + \lambda A - \mu A^{'}=0$

so also,

$b + \lambda B = b^{'} + \mu B^{'}$ or $b-b^{'}+\lambda B - \mu B^{'}=0$

$c + \lambda C = c^{'} + \mu C^{'}$ or $c-c^{'} + \lambda C - \mu C^{'}=0$ .

Eliminating $\lambda$ and $\mu$ , we get the following:

$\left | \begin{array}{ccc} a-a^{'} & A & -A^{'}\\ b-b^{'} & B & -B^{'}\\ c-c^{'} & C & -C^{'} \end{array} \right |=0$ , that is,

$\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0$

Problem II:

A straight line is such that the algebraic sum of the perpendiculars falling upon it from any number of fixed points is zero. Show that it always passes through a fixed point.

Solution II:

Let $(x_{i},y_{i})$ where $i=1,2,3,\ldots$ be n fixed points. Let $ax+by+c=0$ be the given line. Thus, as per given hypthesis, we have

$\sum_{i=1}^{n}\frac{ax{i}+by_{i}+c}{\sqrt{(a^{2}+b^{2})}}=0 \Longrightarrow a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}y_{i}+nc=0 \Longrightarrow a\overline{x}+b\overline{y}+c=0$ where $\overline{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i}$ and $\overline{y}=\frac{}{}\sum_{i=1}^{n}y_{i}$

which shows that the given line passes through the fixed point $(\frac{1}{n}\sum_{i=1}^{n}x_{i}, \frac{1}{n}\sum_{i=1}^{n}y_{i})$ .

Problem III:

The straight lines $L \equiv ax+by+c=0$ and $L_{1} \equiv a_{1}x+b_{1}y+c_{1}=0$ are intersecting. Find the straight line $L_{2}$ such that L is the bisector of the angle between $L_{1}$ and $L_{2}$ .

Solution III:

Let the equation of the line $L_{2}$ be $L_{1}+ \lambda L=0 \Longrightarrow (a_{1}+\lambda a)x+(b_{1}+\lambda b)y+\lambda c=0$ where the slopes of $L_{2}, L, L_{1}$ are respectively

$-\frac{a_{1}+\lambda a}{b_{1}+\lambda b}, \frac{-a}{b}, -\frac{a_{1}}{b_{1}}$ .

Since L is the bisector of the angle between $L_{2}$ and $L_{1}$ we have

$\frac{-(\frac{a_{1}+\lambda a}{b_{1}+\lambda b})+\frac{a}{b}}{1+\frac{a(a_{1}+\lambda a)}{b(b_{1}+\lambda b)}}=\frac{-\frac{a}{b}+\frac{a_{1}}{b_{1}}}{1+\frac{aa_{1}}{bb_{1}}}$

$\Longrightarrow \frac{-b(a_{1}+\lambda a)+a(a_{1}+\lambda b)}{b(b_{1}+\lambda b)+a(a_{1}+\lambda a)}=-\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}$

$\Longrightarrow \frac{ab_{1}-a_{1}b}{\lambda (a^{2}+b^{2})+aa_{1}+bb_{1}} = -\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}$

$\Longrightarrow \lambda = - \frac{2(aa_{1}+bb_{1})}{a^{2}+b^{2}}$

Hence, the equation of the required line $L_{1}$ is $(a^{2}+b^{2})(a_{1}x+b_{1}y+c_{1})=2(aa_{1}+bb_{1})(ax+by+c)$ .

Problem IV:

If a, b are real numbers and $c>0$ , find the locus represented by $|ay-bx|=c\sqrt{(x-a)^{2}+(y-b)^{2}}$ .

PS: Please draw a right angled triangle PMA, with right angle at M, and P being $(x,y)$ and A being $(a,b)$ .

Solution IV:

Let $x=a+r\cos {\theta}$ and $y=b+r\sin {\theta}$ , then the given equation becomes $a\sin {\theta}-b\cos {\theta}=c$ .

$\Longrightarrow r\sin{(\theta-\alpha)}=c$ where $r=\sqrt{a^{2}+b^{2}}$ and $\tan {(\alpha)}=\frac{b}{a}$ which is the slope of $ay-bx$ , which in turn implies $\frac{c}{r}=\sin (\theta -\alpha) \leq 1$

$\Longrightarrow c \leq r$ , or $c \leq \sqrt{a^{2}+b^{2}}$ . The given equation now becomes

$\frac{|ay-bx|}{\sqrt{a^{2}+b^{2}}}=\frac{c}{\sqrt{a^{2}+b^{2}}}\sqrt{(x-a)^{2}+(y-b)^{2}}$ ….call this as relation I.

If M is the foot of the perpendicular from a point P(x,y) on the line $ay-bx=0$ and A is the point $(a,b)$ which clearly lies on this line, then from relation I, we have

$\frac{PM}{PA}=\frac{c}{\sqrt{a^{2}+b^{2}}}=\sin {(\theta - \alpha)}$ . Hence, the locus of P is a straight line through the point $(a,b)$ inclined at an angle $\arcsin {\frac{c}{\sqrt{a^{2}+b^{2}}}}$ with the line $ay-bx=0$ .

Problem V:

Find the co-ordinates of the orthocentre of the triangle formed by the lines $y=0$ and $(1+t)x-ty+t(1+t)=0$ and $(1+u)x-uy+u(1+u)=0$ , where $t \neq u$ , and show that for all values of t and u, the orthocentre lies on the line $x+y=0$ .

Solution V:

Let the equation of the side BC be $y=0$ . Then, the coordinates of B and C are $(-t,0)$ and $(-u,0)$ , respectively, where $(1+t)x-ty+t(1+t)=0$ and $(1+u)x-uy+u(1+u)=0$ are equations of AB and AC, respectively.

PS: Please draw the diagram on your own for a better understanding of the solution presented.

Now, equation of BE is $y={\frac{-u}{1+u}}(x+t)$ …let us call this equaiton I.

And, equation of CF is $y=\frac{-t}{1+t}(x+u)$ …let us call this equation II.

Solving I and II, we get the following:

$x(\frac{u}{1+u}-\frac{t}{1+t})=\frac{tu}{1+t}-\frac{tu}{1+u}$ , which in turn implies that

$x=tu$ and $y=-tu$ , so that the orthocentre is the point $(tu,-tu)$ which lies on the line $x+y=0$ .

Cheers,

Nalin Pithwa

By Nalin Pithwa | Also posted in co-ordinate geometry, IITJEE Advanced Mathematics | Comments (1)

Some random problems/solutions in Coordinate Geometry II: IITJEE mathematics training

February 17, 2018 – 8:45 pm

Question I:

Find the equation of the tangent to the circle $x^{2}-y^{2}-4x-8y+16=0$ at the point $(2+\sqrt{3},3)$ . If the circle rolls up along this tangent by 2 units, find its equation in the new position.

Solution I:

The centre $C_{1}$ of the given circle is $(2,4)$ and its radius is 2. Equation of the tangent at $A(2+\sqrt{3},3)$ to the circle is

$x(2+\sqrt{3})+3y-2(x+2+\sqrt{3})-4(y+3)+16=0$

or $\sqrt{3}x-y-2\sqrt{3}=0$ .

The slope of this line is $\sqrt{3}$ showing that it makes an angle of 60 degrees with the x-axis. After the circle rolls up along the tangent at A through a distance 2 units, its centre moves from $C_{1}$ to $C_{2}$ . We now find the co-ordinates of $C_{2}$ . Since $C_{1}C_{2}$ is parallel to the tangent at A and it passes through $C_{1}$ , $(2,4)$ , its equation is $\frac{x-2}{\cos {\theta}} = \frac{y-4}{\sin {\theta}}$ , where $\theta=60 \deg$ ; $C_{2}$ being at a distance 2 units on this line from $C_{1}$ ; its co-ordinates are

$(2\cos {\theta}+2, 2\sin{\theta}+4)$ , that is, $(3, 4+\sqrt{3})$ .

Hence, the equation of the circle in the new position is

$(x-3)^{2}+(y-(4+\sqrt{3})^{2})=2^{2}$ , which in turn implies that

$x^{2}+y^{2}-6x-2(4+\sqrt{3})y+8(3+\sqrt{3})=0$ .

Question 2:

A triangle has two of its sides along the axes, its third side touches the circle $x^{2}+y^{2}-2ax-2ay+a^{2}=0$ . Prove that the locus of the circumcentre of the triangle is

$a^{2}-2a(x+y)+2xy=0$ .

Solution 2:

The given circle has its centre at $C(a,a)$ and its radius is a so that it touches both the axes along which lie the two sides of the triangle. Let the third side be $\frac{x}{p} + \frac{y}{p}=1$ .

So that A is $(p,a)$ and B is $(a,q)$ and the line AB touches the given circle. Since $\angle AOB$ is a right angle, AB is diameter of the circumcentre of the triangle AOB. So, the circumcentre $P(h,k)$ of the triangle AOB is the mid-point of AB,

that is, $2h=p$ , $2k=q$ .

Now, the equation of AB is $\frac{x}{p} + \frac{y}{q}=1$ , which touches the given circle,

$\frac{a(p+q)-pq}{\sqrt{p^{2}+q^{2}}}=a$

$\Longrightarrow a^{2}(p+q)^{2}+p^{2}-2apq(p+q)=a^{2}(p^{2}+q^{2})$

$\Longrightarrow 2a^{2}-2a(p+q)+pq=0$

$2a^{2}-2a(2h+2k)+2h-2k=0$ .

Hence, the locus of $P(h,k)$ is $a^{2}-2a(x+y)+2xy=0$ .

Question 3:

A circle of radius 2 units rolls on the outerside of the circle $x^{2}+y^{2}+4x=0$ , touching it externally. Find the locus of the centre of this outside circle. Also, find the equations of the common tangents of these two circles when the line joining the centres of the two circles make an angle of 60 degrees with x-axis.

Solution 3:

The centre C of the given circle is $(-2,0)$ and its radius is 2. Let $P(h,k)$ be the centre of the outer circle touching the given circle externally then $CP=2+2=4$ , which in turn implies,

$(h+2)^{2}+k^{2}=4^{2}$

So, the locus of P is $(x+2)^{2}+y^{2}=16$ , or $x^{2}+y^{2}+4x-12=0$ .

Since the two circles touch each other externally,, there are 3 common tangents to these circles.

One will be perpendicular to the line joining the centres and the other two will be parallel to the line joining the centres as the radii of the two circles are equal, co-ordinates of P are given by

$\frac{h+2}{\cos{60 \deg}} = \frac{k-0}{\sin{60 \deg}}=4 \Longrightarrow h=0, k=2\sqrt{3}$ ,

co-ordinates of M, the mid-point of CP is $(-1,\sqrt{3})$ .

Hence, the equation of the common tangent perpendicular to CP is

$y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x+1) \Longrightarrow x+\sqrt{3}y-2=0$ .

Let the equation of the common tangent parallel to CP be $\sqrt{3}x-y+\lambda=0$ .

Since it touches the given circle $\frac{}{}= \pm 2 \Longrightarrow \lambda = 2\sqrt{3} \pm 4$ .

Hence, the other common tangents are $\sqrt{3}x -y \pm 4 + 2\sqrt{3}=0$ .

Question 4:

If $S=0$ and $S^{'}=0$ are the equations of two circles with radii r and $r^{'}$ respectively, then show that the circles $\frac{S}{r} \pm \frac{S^{'}}{r}=0$ cut orthogonally.

Solution 4:

Let the line of centres of the given circle be taken as the x-axis and its mid-point as the origin…Note this is the key simplifying assumption.

If the distance between the centres is 2a, the co-ordinates of the centre are $(a,0)$ and $(-a,0)$ . Hence, we get the following:

$S \equiv (x-a)^{2}+y^{2}-r^{2}=0$ , that is,

$S^{'} \equiv x^{2}+y^{2}-2ax +a^{2}-r^{2}=0$

and $S^{'} \equiv x^{2}+y^{2}+2ax+a^{2}-(r^{'})^{2}=0$ so that $\frac{}{} + \frac{}{}=0 \Longrightarrow Sr^{'}+S^{'}r^{'}=0$ , that is,

$\Longrightarrow (r+r^{'})(x^{2}+y^{2}+a^{2})-2ax(r^{'}-r)-rr^{'}(r+r^{'})=0$

$\Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}-r}{r^{'}+r}x + a^{2}-rr^{'}=0$ …call this I.

and $\frac{S}{r} - \frac{S^{'}}{r^{'}}=0$ and in turn $\Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}+r}{r^{'}-r}+a^{2}+rr^{'}=0$ …call this II.

Now, since $2(\{ -a\frac{r^{'}-r}{r^{'}+r}\})(\{-a\frac{r^{'}+r}{r^{'}-r} \})+ 2\times 0 \times 0 =2a^{2}=(a^{2}-rr^{'}) +(a^{2}+rr^{'})$ .

The circles I and II intersect orthogonally.

Question 5:

Let P, Q, R, S be the centres of the four circles each of which is cut by a fixed circle orthogonally. If $t_{1}^{2}$ , $t_{2}^{2}$ , $t_{3}^{2}$ , $t_{4}^{2}$ be the squares of the lengths of the tangents to the four circles from a point in their plane, then prove that

$t_{1}^{2}\Delta QRS +t_{2}^{2}\Delta RSP + t_{3}^{2}\Delta SPQ + t_{4}^{2}\Delta PQR=0$

Solution 5:

Let the equations of the four circles be

$x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$ , $i=1,2,3,4$ , then centres of these circles are as follows:

$P(-g_{1},-f_{1})$ , $Q(-g_{2},-f_{2})$ , $R(-g_{3},-f_{3})$ , and $S(-g_{4},-f_{4})$

Let the fixed point in the plane be taken as the origin, then $t_{1}^{2}=c_{1}$ , $t_{2}^{2}=c_{2}$ , $t_{3}^{2}=c_{3}$ and $t_{4}^{2}=c_{4}$ . Let the equation of the fixed circle cutting the four circles orthogonally be

$x^{2}+y^{2}+2gx + 2fy +c=0$ , then $2gg_{1}+2ff_{1}=c+c_{1}=c+t_{1}^{2}$ , or

we get the following:

$2gg_{i}+2ff_{i}-c-t_{i}^{2}=0$ , for $i=1,2,3,4$ .

Eliminating the unknowns g, f, c we get

$\left| \begin{array}{cccc} 2g_{1} & 2f_{1} & -1 & -t_{1}^{2} \\ 2g_{2} & 2f_{2} & -1 & -t_{2}^{2}\\ 2g_{3} & 2f_{3} & -1 & -t_{3}^{2}\\ 2g_{4} & 2f_{4} & -1 & -t_{4}^{2} \end{array} \right|$

or, $t_{1}^{2}|D_{1}|-t_{2}^{2}|D_{2}|+t_{3}^{2}|D_{3}|-t_{4}^{2}|D_{4}|=0$

where $|D_{1}|= \left| \begin{array}{ccc} g_{2} & f_{2} & 1\\ g_{1} & f_{1} & 1 \\ g_{4} & f_{4} & 1 \end{array} \right |=2\Delta QRS$ ,

and $|D_{2}|=2\Delta PRS$ , $|D_{3}|=2\Delta PQS$ and $|D_{4}|=2\Delta PQR$

Hence, we get the following:

$t_{1}^{2}\Delta QRS + t_{2}^{2}\Delta RSP + t_{3}^{2}SPQ + t_{4}^{2}PQR=0$ .

Homework Quiz Coordinate Geometry:

OAB is any chord of a circle which passes through O, a point in the plane of the circle and meets it in points A and B. A point P is taken on this chord such that OP is (i) arithmetic mean (ii) geometric mean of OA and OB. Prove that the locus of P in either case is a circle. Determine the circle.
Let $2x^{2}+y^{2}-3xy=0$ be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length OA.
Let P, Q and R be the centres and $r_{1}, r_{2}, r_{3}$ are the radii respectively of three coaxial circles. Show that $r_{1}^{2}QR + r_{2}^{2}RP + r_{3}^{2}PQ=-PQ. QR. RP$
If ABC be any triangle and $A^{'}B^{'}C^{'}$ be the triangle formed by the polars of the points A, B, C with respect to a circle, so that $B^{'}C^{'}$ is the polar of A; $C^{'}A^{'}$ is the polar of B and $A^{'}B^{'}$ is the polar of C. Prove that the lines $AA^{'}$ , $BB^{'}$ and $CC^{'}$ meet in a point.

That’s all, folks !

Nalin Pithwa.

By Nalin Pithwa | Also posted in co-ordinate geometry | Tagged cartesian geometry, coordinate geometry, descartes, IIT JEE mains co ordinate geometry practice, IITJEE Mains maths practice | Comments (0)

Some random sample problems-solutions in Coordinate Geometry: I: IITJEE Mains Maths tutorials

February 16, 2018 – 4:53 pm

Question I:

The point $(4,1)$ undergoes the following transformations, successively:

a) reflection about the line $y=x$ .

b) translation through a distance 2 units along the positive directions of the x-axis.

c) rotation through an angle of $\pi/4$ about the origin in the anticlockwise direction.

d) reflection about $x=0$ .

Hint: draw the diagrams at very step!

Ans: $(1/\sqrt{2}, 7/\sqrt{2})$

Question 2:

$A_{1}, A_{2}, A_{3}, \ldots, A_{n}$ are n points in a plane whose co-ordinates are $(x_{1}, y_{1})$ , $(x_{2},y_{2})$ , $\ldots$ , $(x_{n},y_{n})$ respectively. $A_{1}$ , $A_{2}$ is bisected at the point $G_{1}$ , $G_{1}A_{3}$ is divided in the ratio 1:2 at $G_{2}$ , $G_{2}A_{4}$ is divided in the ratio $1:3$ at $G_{3}$ , $G_{3}A_{3}$ is divided in the ratio $1:4$ at $G_{4}$ and so on until all n points are exhausted. Show that the co-ordinates of the final point so obtained are

$(\frac{1}{n}(x_{1}+x_{2}+ \ldots + x_{n}) , \frac{1}{n}(y_{1}+y_{2}+ \ldots + y_{n}) )$ .

Solution 2:

The co-ordinates of $G_{1}$ are $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$ .

Now, $G_{2}$ divides $G_{1}A_{3}$ in the ratio $1:2$ . Hence, the co-ordinates of $G_{2}$ are

$( \frac{1}{3}(\frac{2(x_{1}+x_{2})}{2}+x_{3}), \frac{1}{3}(\frac{3(y_{1}+y_{2})}{2}+y_{3}))$ , or $(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3})$ .

Again, $G_{3}$ divides $G_{2}A_{4}$ in the ratio $1:4$ . Therefore, the co-ordinates of $G_{3}$ are $(\frac{1}{4}(\frac{3(x_{1}+x_{2}+x_{3})}{3}+x_{4}) ,\frac{1}{4}(\frac{3(y_{1}+y_{2}+y_{3})}{3}+y_{4}) )$ , or

$( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4} )$ .

Proceeding in this manner,we can show that the coordinates of the final point obtained will be

$(\frac{1}{n}(x_{1}+x_{2}+x_{3}+\ldots + x_{n}),\frac{1}{n}(y_{1}+y_{2}+y_{3}+\ldots + y_{n}))$ .

Remark: For a rigorous proof, prove the above by mathematical induction.

Question 3:

A line L intersects the three sides BC, CA, and AB of a triangle ABC at P, Q and R, respectively. Show that $\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1$

Solution 3:

Let $A(x_{1},y_{1})$ , $B(x_{2},y_{2})$ , and $C(x_{3},y_{3})$ be the vertices of $\triangle ABC$ , and let $lx+my+n=0$ be equation of the line L. If P divides BC in the ratio $\lambda:1$ , then the coordinates of P are $(\frac{\lambda x_{3}+x_{2}}{\lambda + 1} ,\frac{\lambda y_{3}+y_{2}}{\lambda + 1})$ .

Also, as P lies on L, we have $l(\frac{\lambda x_{3}+x_{2}}{\lambda + 1})+m(\frac{\lambda y_{3}+y_{2}}{\lambda + 1})+n=0$

$\Longrightarrow \frac{lx_{2}+my_{2}+n}{lx_{3}+my_{3}+n}=\lambda=\frac{BP}{PC}$ …..call this relation I.

Similarly, we can obtain $\frac{CQ}{QA}=-\frac{lx_{3}+my_{3}+n}{lx_{1}+my_{1}+n}$ ….call this relation II.

and so, also, we can prove that $\frac{AR}{RB}=-\frac{lx_{1}+my_{1}+n}{lx_{2}+my_{2}+n}$ …call this III.

Multiplying, I, II and III, we get the desired result.

The above is the famous Menelaus’s theorem of plane geometry proved with elementary tools of co-ordinate geometry. As a homework quiz, try proving the equally famous Ceva’s theorem of plane geometry with elementary tools of co-ordinate geometry.

Question 4:

A triangle has the lines $y=m_{1}x$ and $y=m_{2}x$ as two of its sides, with $m_{1}$ and $m_{2}$ being roots of the equation $bx^{2}+2hx+a=0$ . If $H(a,b)$ is the orthocentre of the triangle, show that the equation of the third side is $(a+b)(ax+by)=ab(a+b-2h)$ .

Solution 4:

Since the given lines intersect at the origin, one of the triangle lies at the origin O(0,0). Let OA and OB be the given lines $y=m_{1}x$ and $y=m_{2}x$ , respectively. Let the equation of AB be $lx+my=1$ . Now, as OH is perpendicular to AB, we have

$\frac{b}{a}=\frac{m}{l}$ , $\Longrightarrow \frac{l}{a}=\frac{m}{b}=k$ , say…call this equation I

Also, the coordinates of A and B are respectively,

$(\frac{1}{l+mm_{1}}, \frac{m_{1}}{l+mm_{1}})$ and $(\frac{1}{l+mm_{2}} , \frac{m_{2}}{l+mm_{2}})$

Therefore, the equation of AB is

$(y-\frac{m_{1}}{l+mm_{1}})=-\frac{1}{m_{2}}(x-\frac{1}{l+mm_{1}})$

or $x+m_{2}y=\frac{1+m_{1}m_{2}}{1+mm_{1}}$ …call this II.

Similarly, the equation of BH is $x+m_{1}y=\frac{1+m_{1}m_{2}}{1+mm_{2}}$ ….call this III.

Solving II and III, we get the coordinates of H. Subtracting III from II, we get

$y=\frac{(1+m_{1}m_{2})m}{l^{2}+lm(m_{1}+m_{2})+m^{2}m_{1}m_{2}}$

Since $m_{1}$ and $m_{2}$ are the roots of the equation $bx^{2}+2hx+a=0$ , we have $m_{1}+m_{2}=-\frac{2h}{b}$ and $m_{1}m_{2}=a/b$ .

$\Longrightarrow y=\frac{(a+b)m}{bl^{2}-2hlm+am^{2}} \Longrightarrow \frac{m}{b}=\frac{bl^{2}-2hlm+am^{2}}{a+b}$ because y=b for H.

$\Longrightarrow k=\frac{k^{2}(ba^{2}-2hab+ab^{2})}{a+b} \Longrightarrow k=\frac{a+b}{ab(a-2h+b)}$ .

Hence, the equation of AB is

$ax+by=\frac{1}{k}=\frac{ab(a+b-2h)}{a+b}$

$\Longrightarrow (a+b)(ax+by)=ab(a+b-2h)$

More later,

Nalin Pithwa.

By Nalin Pithwa | Also posted in co-ordinate geometry | Tagged Ceva's theorem, Menelaus theorem | Comments (0)

Circles and System of Circles: IITJEE Mains: some solved problems I

December 18, 2017 – 8:43 pm

Part I: Multiple Choice Questions:

Example 1:

Locus of the mid-points of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the centre is (a) $x+y=2$ (b) $x^{2}+y^{2}=1$ (c) $x^{2}+y^{2}=2$ (d) $x-y=0$

Answer 1: C.

Solution 1:

Let O be the centre of the circle $x^{2}+y^{2}=4$ , and let AB be any chord of this circle, so that $\angle AOB=\pi /2$ . Let $M(h,x)$ be the mid-point of AB. Then, OM is perpendicular to AB. Hence, $(AB)^{2}=(OA)^{2}+(AM)^{2}=4-2=2 \Longrightarrow h^{2}+k^{2}=2$ . Therefore, the locus of $(h,k)$ is $x^{2}+y^{2}=2$ .

Example 2:

If the equation of one tangent to the circle with centre at $(2,-1)$ from the origin is $3x+y=0$ , then the equation of the other tangent through the origin is (a) $3x-y=0$ (b) $x+3y=0$ (c) $x-3y=0$ (d) $x+2y=0$ .

Answer 2: C.

Solution 2:

Since $3x+y=0$ touches the given circle, its radius equals the length of the perpendicular from the centre $(2,-1)$ to the line $3x+y=0$ . That is,

$r= |\frac{6-1}{\sqrt{9+1}}|=\frac{5}{\sqrt{10}}$ .

Let $y=mx$ be the equation of the other tangent to the circle from the origin. Then,

$|\frac{2m+1}{\sqrt{1+m^{2}}}|=\frac{5}{\sqrt{10}}=25(1+m^{2})=10(2m+1)^{2} \Longrightarrow 3m^{2}+8m-3=0$ , which gives two values of m and hence, the slopes of two tangents from the origin, with the product of the slopes being -1. Since the slope of the given tangent is -3, that of the required tangent is 1/3, and hence, its equation is $x-3y=0$ .

Example 3.

A variable chord is drawn through the origin to the circle $x^{2}+y^{2}-2ax=0$ . The locus of the centre of the circle drawn on this chord as diameter is (a) $x^{2}+y^{2}+ax=0$ (b) $x^{2}+y^{2}+ay=0$ (c) $x^{2}+y^{2}-ax=0$ (d) $x^{2}+ y^{2}-ay=0$ .

Answer c.

Solution 3:

Let $(h,k)$ be the centre of the required circle. Then, $(h,k)$ being the mid-point of the chord of the given circle, its equation is $hx+ky-a(x+h)=h^{2}+k^{2}-2ah$ .

Since it passes through the origin, we have $-ah=h^{2}+k^{2}-2ah \Longrightarrow h^{2}+k^{2}-ah=0$ .

Hence, locus of $(h,k)$ is $x^{2}+y^{2}-ax=0$ .

Quiz problem:

A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is (a) $m(m+n)$ (b) $m+n$ (c) $n(m+n)$ (d) $(1/2)(m+n)$ .

To be continued,

Nalin Pithwa.

By Nalin Pithwa | Also posted in co-ordinate geometry, ISI Kolkatta Entrance Exam, KVPY | Comments (0)

Circles and Systems of Circles: IITJEE mains co-ordinate geometry basics

September 27, 2017 – 1:52 am

Section I:

Definition of a Circle:

A circle is the locus of a point which moves in a plane so that it’s distance from a fixed point in the plane is always constant.The fixed point is called the centre of the circle and the constant distance is called its radius.

Section II:

Equations of a circle:

An equation of a circle with centre $(h,k)$ and radius r is $(x-h)^{2}+(y-k)^{2}=r^{2}$ .
An equation of a circle with centre $(0,0)$ and radius r is $x^{2}+y^{2}=r^{2}$ .
An equation of a circle on the line segment joining $(x_{1},y_{1})$ and $(x_{2},y_{2})$ as diameter is $(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$ .
General equation of a circle is : where g, f, and c are constants
- centre of this circle is $(-g,-f)$
- Its radius is $\sqrt{g^{2}+f^{2}-c}$ , $g^{2}+f^{2}\geq c$
- Length of the intercept made by this circle on the x-axis is $2\sqrt{g^{2}-c}$ if $g^{2}-c \geq 0$ and that on the y-axis is $\sqrt{f^{2}-c}$ if $f^{2}-c \geq 0$ .
General equation of second order degree in x and y represents a circle if and only if:
- coefficient of $x^{2}$ equals coefficient of $y^{2}$ , that is, $a=b \neq 0$
- coefficient of $xy$ is zero, that is , $h=0$
- $g^{2}+f^{2}-ac \geq 0$

Section III: Some results regarding circles:

Position of a point with respect to a circle: Point $P(x_{1},y_{1})$ lies outside, on or inside a circle $S \equiv x^{2}+y^{2}+2gx+2fy+c=0$ , according as $S_{1} \equiv x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c>, =, \hspace{0.1in} or \hspace{0.1in}< 0$
Parametric coordinates of any point on the circle $(x-h)^{2}+(y-k)^{2}=r^{2}$ are given by $(h+r\cos{\theta},k+r\sin{\theta})$ with $0 \leq \theta < 2\pi$ . In particular, parametric coordinates of any point on the circle.
An equation of the tangent to the circle $x^{2}+y^{2}+2gx+2fy+c=0$ at the point $(x_{1},y_{1})$ on the circle is $xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0$
An equation of the normal to the circle $x^{2}+y^{2}+2gx+2fy+c=0$ at the point $(x_{1},y_{1})$ on the circle is $\frac{y-y_{1}}{y_{1}+f} = \frac{x-x_{1}}{x_{1}+g}$
Equations of the tangent and normal to the circle $x^{2}+y^{2}=r^{2}$ at the point $(x_{1},y_{1})$ on the circle are, respectively, $xx_{1}+yy_{1}=r^{2}$ and $\frac{x}{x_{1}}=\frac{y}{y_{1}}$
The line $y=mx+c$ is a tangent to the circle $x^{2}+y^{2}=r^{2}$ if and only if $c^{2}=r^{2}(1+m^{2})$ .
The lines $y=mx \pm r\sqrt{(1+m^{2})}$ are tangents to the circle $x^{2}+y^{2}=r^{2}$ , for all finite values of m. If m is infinite, the tangents are $x \pm r=0$ .
An equation of the chord of the circle $S=x^{2}+y^{2}+2gx+2fy+c=0$ , whose mid-point is $(x_{1},y_{1})$ is $T=S_{1}$ , where $T \equiv xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c$ and $S_{1}=x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c$ . In particular, an equation of the chord of the circle $x^{2}+y^{2}=r^{2}$ , whose mid-point is $(x_{1},y_{1})$ is $xx_{1}+yy_{1}=x_{1}^{2}+y_{1}^{2}$ .
An equation of the chord of contact of the tangents drawn from a point $(x_{1},y_{1})$ outside the circle $S=0$ is $T=0$ .(S and T are as defined in (8) above).
Length of the tangent drawn from a point $(x_{1},y_{1})$ outside the circle $S=0$ , to the circle, is $\sqrt{S_{1}}$ . (S and $\sqrt{S_{1}}$ ) are as defined in (8) above.)
Two circles with centres $C_{1}(x_{1},y_{1})$ and $C_{2}(x_{2},y_{2})$ and radii $r_{1}$ , $r_{2}$ respectively, (i) touch each other externally if $C_{1}C_{2}|=r_{1}+r_{2}$ . the point of contact is $(\frac{r_{1}x_{2}+r_{2}x_{1}}{r_{1}+r_{2}},\frac{r_{1}y_{2}+r_{2}y_{1}}{r_{1}+r_{2}})$ and (ii) touch each other internally if $|C_{1}C_{2}|=|r_{1}-r_{2}|$ , where $r_{1} \neq r_{2}$ ; the point of contact is $(\frac{r_{1}x_{2}-r_{2}x_{1}}{r_{1}-r_{2}} , \frac{r_{1}y_{2}-r_{2}y_{1}}{r_{1}-r_{2}})$
An equation of the family of circles passing through the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) +\lambda(F)=0$ , where

$F=\left|\begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right|$

An equation of the family of circles which touch the line $y-y_{1}=m(x-x_{1})$ at $(x_{1}, y_{1})$ for any finite value of m is $(x-x_{1})^{2}+(y-y_{1})^{2}+\lambda ((y-y_{1})-m(x-x_{1}))=0$ . If m is infinite, the equation becomes $(x-x_{1})^{2}+(y-y_{1})^{2}+\lambda (x-x_{1})=0$ .
Let QR be a chord of a circle passing through the point and let the tangents at the extremities Q and R of this chord intersect at the point . Then, locus of L is called the polar of P with respect to the circle, and P is called the pole of its polar.
- Equation of the polar of $P(x_{1},y_{1})$ with respect to the circle $S \equiv x^{2}+y^{2}+2gx+2fy+c=0$ is $T=0$ , where T is defined as above.
- If the polar of P with respect to a circle passes through Q, then the polar of Q with respect to the same circle passes through P. Two such points P and Q, are called conjugate points of the same circle.
If lengths of the tangents drawn from a point P to the two circles and are equal, then the locus of P is called the radical axis of the two circles and , and its equation is , that is,
- Radical axis of two circles is perpendicular to the line joining their circles.
- Radical axes of three circles, taken in pairs, pass through a fixed point called the radical centre of the three circles, if the centres of these circles are non-collinear.

4: Special Forms of Equation of a Circle:

An equation of a circle with centre $(r,r)$ and radius $|r|$ is $(x-r)^{2}+(y-r)^{2}=r^{2}$ . This touches the co-ordinate axes at the points $(r,0)$ and $(0,r)$ .
An equation of a circle with centre $(x_{1},r)$ , radius $|r|$ is $(x-x_{1})^{2}+(y-r)^{2}=r^{2}$ . This touches the x-axis at $(x_{1},0)$ .
An equation of a circle with centre $(\frac{a}{2},\frac{b}{2})$ and radius $\sqrt{\frac{(a^{2}+b^{2})}{4}}$ is $x^{2}+y^{2}-ax-by=0$ . This circle passes through the origin $(0,0)$ , and has intercepts a and b on the x and y axes, respectively.

5: Systems of Circles:

Let $S \equiv x^{2}+y^{2}+2gx+2fy+c$ ; and $S^{'} \equiv x^{2}+y^{2}+2g^{'}x+2f^{'}y+c^{'}$ and $L \equiv ax + by + k^{'}$ .

If two circles $S=0$ and $S^{'}=0$ intersect at real and distinct points, then $S+\lambda S^{'}=0$ where $\lambda \neq -1$ represents a family of circles passing through these points, where $\lambda$ is a parameter, and $S-S^{'}=0$ when $\lambda=-1$ represents the chord of the circles.
If two circles $S =0$ and $S^{'}=0$ touch each other, then $S-S^{'}=0$ represents equation of the common tangent to the two circles at their point of contact.
If two circles $S=0$ and $S^{'}=0$ intersect each other orthogonally (the tangents at the point of intersection of the two circles are at right angles), then $2gg^{'}+2ff^{'}=c+c^{'}$ .
If the circle $S=0$ intersects the line $L=0$ at two real and distinct points, then $S+\lambda L=0$ represents a family of circles passing through these points.
If $L=0$ is a tangent to the circle $S=0$ at P, then $S+\lambda L=0$ represents a family of circles touching $S=0$ at P, and having $L=0$ as the common tangent at P.
Coaxial Circles: A system of circles is said to be coaxial if every pair of circles of the system have the same radical axis. The simplest form of the equation of a coaxial system of circles is : $x^{2}+y^{2}+2gx+c=0$ , where g is a variable and c is constant, the common radical axis of the system being y-axis and the line of centres being x-axis. The Limiting points of the coaxial system of circles are the members of the system which are of zero radius. Thus, the limiting points of the coaxial system of circles $x^{2}+y^{2}+2gx+c=0$ are $(\pm \sqrt{c},0)$ if $c>0$ . The equation $S+\lambda S^{'}=0$ ( $\lambda \neq -1$ ) represents a family of coaxial circles, two of whose members are given to be $S=0$ and $S^{'}=0$ .
Conjugate systems (or orthogonal systems) of circles : Two system of circles such that every circle of one system cuts every circle of the other system orthogonally are said to be conjugate system of circles. For instance, $x^{2}+y^{2}+2gx+c=0$ and $x^{2}+y^{2}+2fy-c=0$ , where g and f are variables and c is constant, represent two systems of coaxial circles which are conjugate.

6: Common tangents to two circles:

If $(x-g_{1})^{2}+(y-f_{1})^{2}=a_{1}^{2}$ and $(x-g_{2})^{2}+(y-f_{2})^{2}=a_{2}^{2}$ are two circles with centres $C_{1}(g_{1},f_{1})$ and $C_{2}(g_{2},f_{2})$ and radii $a_{1}$ and $a_{2}$ respectively, then we have the following results regarding their common tangents:

When $C_{1}C_{2}>a_{1}+a_{2}$ , that is, distance between the centres is greater than the sum of their radii, the two circles do not intersect with each other, and four common tangents can be drawn to circles. Two of them are direct common tangents and other two are transverse common tangents. The points $T_{1},T_{2}$ of intersection of direct common tangents and transverse common tangents respectively, always lie on the line joining the centres of the two circles and divide it externally and internally respectively in the ratio of their radii.
When $C_{1}C_{2}=a_{1}+a_{2}$ , that is, the distance between the centres is equal to the sum of their radii, the two circles touch each other externally, two direct tangents are real and distinct and the transverse tangents coincide.
When $C_{1}C_{2}<a_{1}+a_{2}$ , that is, the distance between the centres is less than the sum of the radii, the circles intersect at two real and distinct points, the two direct common tangents are real and distinct while the transverse common tangents are imaginary.
When $C_{1}C_{2}=|a_{1}-a_{2}|$ with $a_{1} \neq a_{2}$ , that is, the distance between the centres is equal to the difference of their radii, the circles touch each other internally, two direct common tangents are real and coincident, while the transeverse common tangents are imaginary.
When $C_{1}C_{2}<|a_{1}-a_{2}|$ , with $a_{1} \neq a_{2}$ , that is, the distance between the centres is less than the difference of the radii, one circle with smaller radius lies inside the other and the four common tangents are all imaginary.

To be continued,

Nalin Pithwa.

By Nalin Pithwa | Also posted in co-ordinate geometry, KVPY | Comments (0)

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