Category Archives: IITJEE Mains

Theory of Equations: Part I: IITJEE maths

Algebra for IITJEE Mains: harder factors: Mathematics Hothouse

Derivatives: IITJEE Mains Maths: Mathematics Hothouse video lecture

Limits Part 2: IITJEE Mains maths: Mathematics Hothouse

Limits part 1: video lecture: IITJEE Mains maths: Mathematics Hothouse

 

Applications of Derivatives IITJEE Maths tutorial: practice problems part IV

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.

Applications of Derivatives: Tutorial: IITJEE Maths: Part II

Another set of “easy to moderately difficult” questions:

  1. 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?
  2. 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?
  3. 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?
  4. 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).
  5. 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).
  6. In which interval does the function y=\frac{x}{\log(x)} increase?
  7. Which is the larger of the functions \sin(x) + \tan(x) and f(x)=2x in the interval (0<x<\pi/2)?
  8. Find the set of all x for which \log {(1+x)} \leq x.
  9. 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 ?
  10. There are exactly two distinct linear functions (find them), such that they map [-1,1] and [0,2].

more later, cheers,

Nalin Pithwa.

Applications of Derivatives: A Quick Review

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.

Co-ordinate geometry practice for IITJEE Maths: Ellipses

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

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

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

(c) x^{2}+y^{2}-5ax+6a^{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.