co-ordinate geometry « Mathematics Hothouse

Category Archives: co-ordinate geometry

Follow Descartes’ Historically Famous Problems !

May 24, 2017 – 1:59 am

Problem 1:

Three circles touching one another externally have radii $r_{1}$ , $r_{2}$ and $r_{3}$ . Determine the radii of the two circles that can be drawn touching all the three circles.

Problem 2:

Consider a circle, say (numbered 1) of unit radius 1. Inside this circle, two circles are drawn (say, numbered 2 and 3), each of radius $\frac{1}{2}$ , which touch each other externally and the first circle internally. Determine the radius of the fourth circle, which touches circles 2 and 3 externally and circle 1 internally. Determine the radius of the fifth circle, which touches each of the circles 2, 3, and 4 externally. Determine the radius of the sixth circle, which touches circles 2 and 4 externally and circle 1 internally. One might notice that curvature of all such circles drawn within the first circle has integer curvature!

It is such historically famous problems (within scope of IITJEE Mains and IITJEE Advanced Maths) which all students should try to internalize all the concepts of Math for IITJEE. Also, in a similar vein, you should practice deriving all basic formulae, relationships of co-ordinate geometry.

More later,

Nalin Pithwa.

Conic Sections (Parabola, Ellipse, Hyperbola): some basic theorems/questions: IITJEE Mains Maths

April 3, 2017 – 1:12 am

Problem 1:

(A) Proposition 1: Show that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.

Proof 1:

Let $t_{1}$ and $t_{2}$ be the extremities of a focal chord of the parabola $y^{2}=4ax$ . Then, it can be shown that $t_{1}t_{2}=-1$ . (Try this part on your own and let me know; if you can’t produce the proof, I will send it to you.) The equations of the tangents at $t_{1}$ and $t_{2}$ are $t_{1}y=x+at_{1}^{2}$ and $t_{2}y=x+at_{2}^{2}$ . The product of the slopes is $\frac{1}{t_{1}}-\frac{1}{t_{2}}=-1$

Therefore, the tangents are at right angles. Also, the point of intersection of these tangents is $x=at_{1}t_{2}$ , $y=a(t_{1}+t_{2})$ , that is, $x=-a$ , $y=a(t_{1}+t_{2})$ , which clearly lies on the directrix $x=-a$ .

(B) Proposition 2: The tangent at any point of a parabola bisects the angle between the focal distance of the point and the perpendicular on the directrix from that point. Homework !

(C) Proposition 3: The portion of a tangent to a parabola cut off between the directrix and the curve, subtends a right angle at the focus. Homework!

Problem 2:

Theorem:

Show that in general, three normals can be drawn to a given parabola from a given point, one of which is always real. Also, show that the sum of the ordinates of the feet of these co-normal points is zero.

Proof:

Let $y^{2}=4ax$ be the given parabola. The equation of a normal to this parabola at $(am^{2},-2am)$ is $y=mx-2am-am^{2}$ . If it passes through a given point $(\alpha, \beta)$ then

$\beta = m\alpha -2am-am^{3}$ , or $am^{3}+(2a-\alpha)m + \beta = 0$ …call this Equation (I)

which, being cubic in m, gives three values of m, say $m_{1}$ . $m_{2}$ , and $m_{3}$ , and hence, three points on the parabola, the normals at which pass through $(\alpha, \beta)$ . Since the complex roots of the equation with real coefficients occur in pairs and the degree of the above equation is odd, at least one of the roots is real so there is at least one real normal to the parabola passing through the given point $(\alpha, \beta)$ .

From (I), we have

$m_{1}+m_{2}+m_{3}=0 \Longrightarrow -2am_{1}-2am_{2}-2am_{3}=0 \Longrightarrow y_{1}+y_{2}+y_{3}=0$ , where $y_{i}=-2am_{i}$ for $i=1, 2, 3$ . Hence, the result.

Problem 3:

If the tangents and normals at the extremities of a focal chord of the parabola $y^{2}=4ax$ intersect at $(x_{1},y_{1})$ and $(x_{2},y_{2})$ respectively, then show that $y_{1}=y_{2}$ .

Answer 3:

Let $P(at_{1}^{2}, 2at_{1})$ and $Q(at_{2}^{2},2at_{2})$ be the extremities of a focal chord of the parabola $y^{2}=4ax$ , then

$t_{1}t_{2}=-1$ ….call this Relation (I).

Next, equations of the tangents at P and Q are respectively, $t_{1}y=x+at_{1}^{2}$ and $t_{2}y=x+at_{2}^{2}$

Solving these equations, we get

$(t_{1}-t_{2})y=a(t_{1}^{2}-t_{2}^{2}) \Longrightarrow y=a(t_{1}+t_{2})$ and

$x=t_{1}a(t_{1}+t_{2})-at_{1}^{2}=at_{1}t_{2}=-a$

So that $x_{1}=-a$ and $y_{1}=a(t_{1}+t_{2})$ ….call this Relation (II).

Now, equations of the normals at P and Q are respectively

$y = -t_{1}x+2at_{1}+at_{1}^{3}$ and $y=-t_{2}x+2at_{2}+at_{2}^{3}$

Solving these equations we get

$-(t_{1}-t_{2})x+2a(t_{1}-t_{2})+a(t_{1}^{3}-t_{2}^{3})=0$

$\Longrightarrow x =2a +a(t_{1}^{2}+t_{2}^{2}+t_{1}t_{2}) = a(t_{1}^{2}+t_{2}^{2}+1)$ (using relation I)

and $y = -t_{1}a(t_{1}^{2}+t_{2}^{2}+1) + 2at_{1}+at_{1}^{3} = -at_{1}t_{2}^{2}+at_{1}=a(t_{1}+t_{2})$ (again by using relationship I),

So that $x_{2}=a(t_{1}^{2}+t_{2}^{2}+1)$ and $y_{2}=a(t_{1}+t_{2})$ so that we get $y_{1}=y_{2}$ .

Note:

Results or relations I and II should be memorized as they are frequently used in the theory and applications.

More later,

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains | Comments (0)

Practice Quiz on Conic Sections (Parabola, Ellipse, Hyperbola): IITJEE Mains — basics 1

April 2, 2017 – 1:12 pm

Multiple Choice Questions:

Problem 1:

A line bisecting the ordinate PN of a point P $(at^{2}, 2at)$ , $t>0$ , on the parabola $y^{2}=4ax$ is drawn parallel to the axis to meet the curve at Q. If NQ meets the tangent at the vertex at the point T, then the coordinates of T are

(a) $(0, \frac{4}{3}at)$ (b) Slatex (0,2at)$ (c) $(\frac{1}{4}at^{2},at)$ (d) $(0,at)$

Problem 2:

If P, Q, R are three points on a parabola $y^{2}=4ax$ whose ordinates are in geometrical progression, then the tangents at P and R meet on

(a) the line through Q parallel to x-axis

(b) the line through Q parallel to y-axis

(d) the line through Q to the focus.

Problem 3:

The locus of the midpoint of the line segment joining the focus to a moving point on the parabola $y^{2}=4ax$ is another parabola with directrix:

(a) $x=-a$ (b) $x=-a/2$ (c) $x=0$ (d) $x=a/2$

Problem 4:

Equation of the locus of the pole with respect to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ , of any tangent line to the auxiliary circle is the curve $\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}}=\lambda^{2}$ where

(a) $\lambda^{2}=a^{2}$ (b) $\lambda^{2}=\frac{1}{a^{2}}$ (c) $\lambda^{2}=b^{2}$ (d) $\lambda^{2}=\frac{1}{b^{2}}$

Problem 5:

The locus of the points of the intersection of the tangents at the extremities of the chords of the ellipse $x^{2}+2y^{2}=6$ , which touch the ellipse $x^{2}+4y^{2}=4$ is

(a) $x^{2}+y^{2}=4$ (b) $x^{2}+y^{2}=6$ (c) $x^{2}+y^{2}=9$ (d) none of these.

Problem 6:

If an ellipse slides between two perpendicular straight lines, then the locus of its centre is

(a) a parabola (b) an ellipse (c) a hyperbola (d) a circle

Problem 7:

Let $P(a\sec{\theta}, b\tan{\theta})$ and $Q(a\sec{\phi}, b\tan{\phi})$ where $\theta + \phi=\pi/2$ , be two points on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$ . If $(h,k)$ is the point of intersection of normals at P and Q, then k is equal to

(a) $\frac{a^{2}+b^{2}}{a}$ (b) $-(\frac{a^{2}+b^{2}}{a})$ (c) $\frac{a^{2}+b^{2}}{b}$ (d) $-(\frac{a^{2}+b^{2}}{b})$

Problem 8:

A straight line touches the rectangular hyperbola $9x^{2}-9y^{2}=8$ , and the parabola $y^{2}=32x$ . An equation of the line is

(a) $9x+3y-8=0$ (b) $9x-3y+8=0$ (c) $9x+3y+8=0$ (d) $9x-3y-8=0$

There could be multiple answers to this question.

Problem 9:

Two parabolas C and D intersect at the two different points, where C is $y=x^{2}-3$ and D is $y=kx^{2}$ . The intersection at which the x-value is positive is designated point A, and $x=a$ at this intersection. The tangent line l at A to the curve D intersects curve C at point B, other than A. If x-value of point B is 1(one), then what is a equal to?

Problem 10:

The triangle formed by the tangent to the parabola $y=x^{2}$ at the point whose abscissa is $x_{0}(x_{0} \in [1,2])$ , the y-axis and the straight line $y=a_{0}^{2}$ has the greatest area if $x_{0}= ?$ . Fill in the question mark!

More later,

Nalin Pithwa.

PS: I think let’s continue this focus on co-ordinate geometry of IITJEE Mains Maths for some more time.

By nkpithwa | Also posted in IITJEE Mains | Comments (0)

Some more examples of pair of straight lines questions for IITJEE math

March 13, 2017 – 11:07 am

Problem 1:

If one of the straight lines given by the equation $ax^{2}+2hxy+by^{2}=0$ coincides with one of those given by $a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0$ , and the other lines represented by them be perpendicular, prove that $\frac{ha^{'}b^{'}}{b^{'}-a^{'}}=\frac{h^{'}ab}{b-a}=\frac{1}{2}\sqrt{aa^{'}bb^{'}}$

Solution 1:

Let the lines represented by $ax^{2}+2hxy+by^{2}=0$ be $y=m_{1}x$ and $y=m_{2}x$ , so that

$m_{1}+m_{2}=\frac{2h}{b}$ and $m_{1}m_{2}=\frac{a}{b}$ …call this equation (A)

The lines represented by $a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0$ are $y=m_{1}x$ and $y=-\frac{1}{m_{2}}x_{1}$ so that

$m_{1}-\frac{1}{m_{2}}=-\frac{2h^{'}}{b^{'}}$ and $m_{1}(-\frac{1}{m_{2}})=\frac{a^{'}}{b^{'}}$ …call this equation (B)

From (A) and (B), we get $m_{1}m_{2}(-\frac{m_{1}}{m_{2}})=\frac{aa^{'}}{bb^{'}}$

which in turn $\Longrightarrow m_{1}^{2}=\frac{aa^{'}}{bb^{'}}$ ,

which in turn $\Longrightarrow m_{1}=\sqrt{(-\frac{aa^{'}}{bb^{'}})}$

and from (A), again, we get $\sqrt{-\frac{aa^{'}}{bb^{'}}}-\sqrt{-\frac{aa^{'}}{ba^{'}}}=-\frac{2h}{b}$

and this in turn $\Longrightarrow \sqrt{(-aa^{'}bb^{'})}(\frac{1}{bb^{'}}-\frac{1}{a^{'}b})=-\frac{2h}{b}$ , that is, it $\Longrightarrow \frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{ha^{'}b^{'}}{b^{'}-a^{'}}$ .

Similarly, from (B), we get $\frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{h^{'}ab}{b-a}$ . Hence, the required result follows.

Problem 2:

If the equation $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represents two straight lines, prove that the equation of the third pair of straight lines passing through the points where these meet the axes $xy=0$ is $c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0$ .

Solution 2:

Equation of the lines passing through the intersection of the given lines and axes is

$ax^{2}+2hxy+by^{2}+2gx+2fy+c+2xy \lambda =0$ , or

$ax^{2}+2(h+\lambda)xy+by^{2}+2gx+2fy+c=0$

Since it represents a pair of straight lines

$abc+2fg(h+\lambda)-af^{2}-bg^{2}-c(h+\lambda)^{2}=0$ ,

$\Longrightarrow abc+2fgh-af^{2}-bg^{2}-ch^{2}+2fg\lambda -2ch\lambda -c\lambda^{2}=0$

$\Longrightarrow \lambda = \frac{2(fg-ch)}{c}$ (since $abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$ ).

Hence, the required equation of the lines is $c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0$

Problem 3:

Show that the equation $\sqrt{3}x^{2}-(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}-\sqrt{3}y^{2}=0$ represents three straight lines through the origin such that one of them bisects the angle between the other two. Also find the equation of the lines perpendicular to the given lines through the origin.

Solution 3:

The given equation can be written as $\sqrt{3}(x^{3}-y^{3})-(4+\sqrt{3})xy(x-y)=0$ , or

$(x-y)(\sqrt{3}(x^{2}+xy+y^{2}))-(4+\sqrt{3})xy=0$ , or

$(x-y)(\sqrt{3}x^{2}-4xy+\sqrt{3}y^{2})=0$ , or

$(x-y)(\sqrt{3}x-y)(x-\sqrt{3}y)=0$ ,or

$x-y=0$ , $\sqrt{3}x-y=0$ , and $x-\sqrt{3}y=0$

which gives three straight lines passing through the origin with slopes 45 degrees, 60 degrees, and 30 degrees respectively showing that $x-y=0$ bisects the angles between the other two.

Now, equations of the lines through the origin perpendicular to these lines are

$x+y=0$ , $x+\sqrt{3}y=0$ , $\sqrt{3}x+y=0$ so their joint equation is given by

$(x+y)(x+\sqrt{3}y)(\sqrt{3}x+y)=0$

or, $\sqrt{3}x^{3}+(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}+\sqrt{3}y^{3}=0$

That’s all, folks !

Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains | Comments (0)

Some solved examples based on pairs of straight lines

March 11, 2017 – 9:07 am

The following questions use the previous (previous blog)theory, concepts and formulae related to pair of straight lines:

Problem 1:

Prove that the straight lines joining the origin to the points of intersection of the straight line $hx + ky =2hk$ and the curve $(x-k)^{2}+(y-h)^{2}=c^{2}$ are at right angles if $h^{2}+k^{2}=c^{2}$ .

Solution 1:

Making the equation of the curve homogeneous with the help of the equation of the line, we get

$x^{2}+y^{2} - 2(kx+hy)(\frac{hx+ky}{2hk})+(h^{2}+k^{2}-c^{2})(\frac{hx+ky}{2hk})^{2}=0$ , or

$4h^{2}k^{2}x^{2} + 4h^{2}k^{2}y^{2}-4hk^{2}x(hx+ky) - 4h^{2}ky(hx+ky)+(h^{2}+k^{2}-c^{2})(h^{2}x^{2}+k^{2}y^{2}+2hxy)=0$

This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve. They will be at right angles if

coefficient of $x^{2}$ + coefficient of $y^{2}$ is zero, that is,

$\Longrightarrow (h^{2}+k^{2})(h^{2}+k^{2}-c^{2})=0$

$\Longrightarrow h^{2}+k^{2}=c^{2}$ (since $h^{2}+k^{2} \neq 0$ ).

Problem 2:

Prove that the two straight lines $(x^{2}+y^{2})(\cos^{2}{\theta}\sin^{2}{\alpha}+\sin^{2}{\theta})=(x\tan{\alpha}+y\sin{\theta})^{2}$ include an angle $2\alpha$ .

Solution 2:

The given equation can be written as

$(\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta} - \tan^{2}{\alpha})x^{2} + 2\tan{\alpha}\sin{\theta}xy + \cos^{2}{\theta}\sin^{2}{\alpha}y^{2}=0$

Here, $a=\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta}-\tan^{2}{\alpha}$ ,

$b=\cos^{2}{\theta}\sin^{2}{\alpha}$ and $h=\tan{\alpha}\sin{\theta}$ .

We need to show that

$2\alpha = \arctan {2\frac{\sqrt{h^{2}-ab}}{a+b}}$

or $\tan{2\alpha} = \frac{2\sqrt{h^{2}-ab}}{a+b}$

Now, $\frac{2\sqrt{h^{2}-ab}}{a+b}$

$= \frac{2\sqrt{\tan^{2}{\alpha}\sin^{2}{\theta}-(\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta}-\tan^{2}{\alpha})\cos^{2}{\theta}\sin^{2}{\alpha}}}{2\cos^{2}{\theta}\sin^{2}{\alpha}+\sin^{2}{\theta}-\tan^{2}{\alpha}}$

$= \frac{2\tan{\alpha}\sqrt{\sin^{2}{\theta}-\cos^{4}{\theta}\sin^{2}{\alpha}\cos^{2}{\alpha}-\sin^{2}{\theta}\cos^{2}{\theta}\cos^{2}{\alpha}+\sin^{2}{\alpha}\cos^{2}{\theta}}}{2\cos^{2}{\theta}\sin^{2}{\alpha}+1-\cos^{2}{\theta}-\tan^{2}{\alpha}}$

$= \frac{2\tan{\alpha}\sqrt{1-\cos^{2}{\theta}-\cos^{4}{\theta}{\cos^{2}{\alpha(1-\cos^{2}{\alpha})-(1-\cos^{2}{\theta})\cos^{2}{\theta}\cos^{2}{\alpha}+(1-\cos^{2}{\alpha})\cos^{2}{\theta}}}}}{1-\tan^{2}{\alpha}-\cos^{2}{\theta}(1-2\sin^{2}{\alpha})}$

$=\frac{2\tan{\alpha}\sqrt{1+\cos^{4}{\theta}\cos^{4}{\alpha}-2\cos^{2}{\theta}\cos^{2}{\alpha}}}{1-\tan^{2}{\alpha}-\cos^{2}{\theta}(\cos^{2}{\theta}-\sin^{2}{\alpha})}$

$= \frac{2\tan{\alpha}(1-\cos^{2}{\theta}\cos^{2}{\alpha})}{(1-\tan^{2}{\alpha})(1-\cos^{2}{\theta}\cos^{2}{\alpha})}=\tan{2\alpha}$

Hence, the given lines include an angle $2\alpha$ .

More later,

Nalin Pithwa.

By nkpithwa | Comments (0)

Pair of straight lines — quick review for IITJEE Math

March 8, 2017 – 9:30 pm

Pair of Straight Lines:

I: Equation of Family of Lines:

A first degree equation $ax+by+c=0$ represents a straight line involving three constants a, b and c, which can be reduced to two by dividing both the sides of the equation by a non-zero constant. For instance, if $a \neq 0$ we can write the equation as

$x+\frac{b}{a}y+\frac{c}{a}=0$ or $x+By+C=0$

So, in order to determine an equation of a line, we need two conditions on the line to determine these constants. For instance, if we know two points on the line or a point on the line and its slope etc., we know the line. But, if we know just one condition, we have infinite number of lines satisfying the given condition. In this case, the equation of the line contains an arbitrary constant and for different values of the constant we have different lines satisfying the given condition and the constant is called a parameter.

Some Equations of Family of Lines:

Family of lines with given slope. (Family of given lines) $y = mx +k$ , where k is a parameter represents a family of lines in which each line has slope m.
Family of lines through a point: $y-y_{0}=k(x-x_{0})$ , where k is a parameter represents a family of lines in which each line passes through the point $(x_{0},y_{0})$ .
Family of lines parallel to a given line: $ax + by + k=0$ , where k is a parameter represents a family of lines which are parallel to the line $ax+by+c=0$ .
Family of lines through intersection of two given lines:
$a_{1}x+b_{1}y+c_{1}+k(a_{2}x+b_{2}y+c_{2})=0$ , where k is a parameter represents a family of lines, in which each line passes through the point of intersection of two intersecting llines $a_{1}x+b_{1}y+c_{1}=0$ and $a_{2}x+b_{2}y+c_{2}=0$ .
Family of lines perpendicular to a given line:
$bx-ay+c=0$ where k is a parameter represents a family of lines, in which each line is perpendicular to the line $ax+by+c=0$ .
Family of lines making a given intercept on axes:
(a) $\frac{x}{a}+\frac{y}{k}=1$ , where k is a parameter, represents a family of lines, in which each line makes an intercept a on the axis of x. and,

6(b) $\frac{x}{k} + \frac{y}{b}=1$ , where k is a parameter, represents a family of lines, in which each line makes an intercept b on the axis of y.

7. Family of lines at a constant distance from the origin:

$x\cos {\alpha}+y\sin{\alpha}=p$ , where $\alpha$ is a parameter, represents a family of lines, in which each line is at a distance p from the origin.

II: Pair of Straight Lines:

A) Pair of straight lines through the origin: A homogeneous equation of the second degree $ax^{2}+2hxy+by^{2}=0$ …call this equation 1; represents a pair of straight lines passing through the origin if and only if $h^{2}-ab \geq 0$ . If $b \neq 0$ , the given equation can be written as

$y^{2}+\frac{2h}{b}xy+\frac{a}{b}x^{2}=0$ , or $(\frac{y}{x})^{2}+\frac{2h}{b}(\frac{y}{x})+\frac{a}{b}=0$ , which, being quadratic in $y/x$ , gives two values of $y/x$ , say $m_{1}$ and $m_{2}$ and hence, the equations $y=m_{1}x$ and $y=m_{2}x$ of two straight lines passing through the origin. The slopes $m_{1}$ and $m_{2}$ of these straight lines are given by the relations

$m_{1}+m_{2}=-\frac{2h}{b}$ …call this equation (2).

$m_{1}m_{2}=\frac{a}{b}$ …call this equation (3).

B) Angle between the lines: represented by $ax^{2}+2hxy+by^{2}=0$ . If $\theta$ is an angle, between the lines represented by (1), then

$\tan{\theta} = \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}=\frac{\pm\sqrt{(m_{1}+m_{2})^{2}-4m_{1}m_{2}}}{1+m_{1}m_{2}} = \frac{\pm \sqrt{4h^{2}-4ab}}{a+b}$

$\Longrightarrow \theta = \arctan (\frac{\pm 2\sqrt{h^{2}-ab}}{a+b})$

If $(a+b)=0$ , the lines are perpendicular, and if $h^{2}=ab$ , then the lines are coincident.

C) Equation of the bisectors of the angles between the lines $ax^{2}+2hxy=by^{2}=0$ .

Let $y=m_{1}x$ and $y=m_{2}x$ be the equation of the straight lines represented by the given equation. Then, equations of the angle bisectors are:

$\frac{y-m_{1}x}{\sqrt{1+m_{1}^{2}}} = \frac{y-m_{2}x}{\sqrt{1+m_{1}^{2}}}$

The joint equation of these bisectors can be written as

$(\frac{y-m_{1}x}{\sqrt{1+m_{1}^{2}}} + \frac{y-m_{2}x}{1+m_{2}^{2}})(\frac{y-m_{1}x}{1+m_{1}^{2}} - \frac{y-m_{2}x}{1+m_{2}^{2}})=0$

or, $(1+m_{2}^{2})(y-m_{1}x)^{2} - (1+m_{1}^{2})(y-m_{2}x)^{2}=0$

or, $x^{2}(m_{1}^{2}-m_{2}^{2}) - 2xy(m_{1}-m_{2})(1-m_{1}m_{2}) + y^{2}(m_{2}^{2}-m_{1}^{2})=0$

or, $\frac{x^{2}-y^{2}}{2xy} = \frac{1-m_{1}m_{2}}{m_{1}+m_{2}} = \frac{a-b}{2h}$ ….this comes from (2) and (3).

or, $\frac{x^{2}-y^{2}}{a-b} = \frac{xy}{h}$

D) The general equation of second degree:

$ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represents a pair of straight lines if and only if

$abc + 2fgh - af^{2} -bg^{2} - ch^{2} =0$ , that is, if

$\left | \begin{array}{ccc} a & h & g \\ h & b & f \\ g & f & c \end{array} \right |=0$

If $h^{2}=ab$ , the co-ordinates $(x_{1},y_{1})$ of the point of intersection of these lines are obtained by solving the equations

$ax_{1}+hy_{1}+x=0$ and $hx_{1}+by_{1}+f=0$

$\Longrightarrow x_{1}=\frac{hf-bg}{ab-h^{2}}$ and $y_{1}=\frac{gh-af}{ab-h^{2}}$

E) Equation $ax^{2}+2hxy +by^{2}=0$ represents a pair of straight lines through the origin parallel to the lines given by the general equation of second degree given in (4) and hence, the angle between the lines given in (4) is same as in (2), that is,

$\theta = \arctan {\frac{\pm {h^{2}-ab}}{a+b}}$

so that if $a+b=0$ , the lines are perpendicular and if $h^{2}=ab$ , the lines are parallel.

F) Equation of the lines joining the origin to the points of intersection of a line and a conic:

Let $L \equiv lx + my + n=0$

and $S \equiv ax^{2} + 2hxy + by^{2} + 2gx + 2fy + c=0$

be the equations of a line and a conic, respectively. Writing the equation of the line as

$\frac{lx + my}{-n} = 1$ and making $S=0$ homogeneous with its help, we get $S=ax^{2}+2hxy+by^{2}+2(gx +fy)$

$(\frac{lx+my}{-n}) + c(\frac{lx+my}{-n})^{2}=0$ , which being a homogeneous equation of second degree, represents a pair of straight lines through the origin and passing through the points common to $S=0$ and $L=0$ .

G) Some important results regarding pair of lines:

(i) Equation of the pair of lines through the origin perpendicular to the pair of lines $ax^{2}+2hxy+by^{2}=0$ is $bx^{2}+2hxy +ay^{2}=0$ .

(ii) The product of the perpendicular lines drawn from the point $(x_{1}, y_{1})$ on the lines $ax^{2}+2hxy+by^{2}=0$ is $|\frac{ax_{1}^{2}+2hx_{1}y_{1}+by_{1}^{2}}{\sqrt{(a-b)^{2}+4h^{2}}}|$

(iii) The product of the perpendiculars drawn from the origin to the lines $ax^{2}+2hxy + by^{2}+2gx+2fy+c=0$ is $\frac{c}{\sqrt{(a-b)^{2}+4h^{2}}}$

(iv) Equations of the bisectors of the angles between the lines represented by $ax^{2}+2hxy + by^{2} +2gx + 2fy + c=0$ are given by

$\frac{(x-x_{1})^{2}-(y-y_{1})^{2}}{a-b} = \frac{(x-x_{1})(y-y_{1})}{h}$ where $(x_{1}, y_{1})$ is the point of intersection of the lines represented by the given equation.

(v) If $ax^{2}+2hxy + by^{2}+2gx + 2fy + c=0$ represents two parallel straight lines, then the distance between them is $2\sqrt{\frac{g^{2}-ac}{a(a+b)}}$

Every nth degree homogeneous equation $a_{0}x^{n}+a_{1}x^{n-1}y + a_{2}x^{n-2}y + \ldots + a_{n-1}xy^{n-1}+a_{n}y^{n}=0$ , then

$m_{1}+m_{2}+ \ldots + m_{n} = - \frac{a_{n-1}}{a_{n}}$

and $m_{1}m_{2}\ldots m_{n}=(-1)^{n}\frac{a_{0}}{a_{n}}$

To be continued later,

-Nalin Pithwa.

By nkpithwa | Also posted in IITJEE Mains, KVPY | Comments (0)

Equation of a circle

March 21, 2016 – 1:04 pm

Consider a fixed complex number $z_{0}$ and let $z$ be any complex number which moves in such a way that its distance from $z_{0}$ is always equal to r. This implies $z$ would lie on a circle whose centre is $z_{0}$ and radius r. And, its equation would be

$|z-z_{0}|=r$

or $|z-z_{0}|^{2}=r^{2}$

or $(z-z_{0})(\overline{z}-\overline{z_{0}})=r^{2}$ ,

or $z\overline{z}-z \overline{z_{0}}-\overline{z}z_{0}-r^{2}=0$

Let $-a=z_{0}$ and $z_{0}\overline{z_{0}}-r^{2}=b$ . Then,

$z\overline{z}+a\overline{z}+\overline{a}z+b=0$

It represents the general equation of a circle in the complex plane.

Now, let us consider a circle described on a line segment AB $(A(z_{1}), B(z_{2}))$ as its diameter. Let $P(z)$ as its diameter. Let $P(z)$ be any point on the circle. As the angle in the semicircle is $\pi/2$ , so

$\angle {APB}=\pi/2$

$\Longrightarrow (\frac{z_{1}-z}{z_{2}-z})=\pm \pi/2$

$\Longrightarrow \frac{z-z_{1}}{z-z_{2}}$ is purely imaginary.

$\frac{z-z_{1}}{z-z_{2}}+\frac{\overline{z}-\overline{z_{1}}}{\overline{z}-\overline{z_{2}}}=0$

$\Longrightarrow (z-z_{1})(\overline{z}-\overline{z_{2}})+(z-z_{2})(\overline{z}-\overline{z_{1}})=0$

Condition for four points to be concyclic:

Let ABCD be a cyclic quadrilateral such that $A(z_{1})$ , $B(z_{2})$ , $C(z_{3})$ and $D(z_{4})$ lie on a circle. (Remember the following basic property of concyclic quadrilaterals: opposite angles are supplementary).

The above property means the following:

$\arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})+\arg (\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$

$\Longrightarrow \arg (\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})=\pi$

$(\frac{z_{4}-z_{1}}{z_{2}-z_{1}})(\frac{z_{2}-z_{3}}{z_{4}-z_{3}})$ is purely real.

Thus, points $A(z_{1})$ , $B(z_{2})$ , $C(z_{3})$ , $D(z_{4})$ (taken in order) would be concyclic if the above condition is satisfied.

More later,

Nalin Pithwa

By nkpithwa | Also posted in Complex Numbers | Tagged circle equation, concyclic points | Comments (0)

Equation of a line : Geometry and Complex Numbers

March 10, 2016 – 3:31 am

Equation of the line passing through the point $z_{1}$ and $z_{2}$ :

Ref: Mathematics for Joint Entrance Examination JEE (Advanced), Second Edition, Algebra, G Tewani.

There are two forms of this equation, as given below:

$\left | \begin{array}{ccc} z & \overline{z_{1}} & 1 \\ z_{1} & \overline{z_{2}} & 1 \\ z_{2} & \overline{z_{2}} & 1 \end{array} \right |=0$

and $\frac{z-z_{1}}{\overline{z}-\overline{z_{1}}}=\frac{z_{1}-z_{2}}{\overline{z_{}}-\overline{z_{2}}}$

Proof:

Let $z_{1}=x_{1}+iy_{1}$ and $z_{2}=x_{2}+iy_{2}$ . Let A and B be the points representing $z_{1}$ and $z_{2}$ respectively.

Let $P(z)$ be any point on the line joining A and B. Let $z=x+iy$ . Then $P \equiv (x,y)$ , $A \equiv (x_{1}, y_{1})$ and $B \equiv (x_{2},y_{2})$ . Points P, A, and B are collinear.

See attached JPEG figure 1.

The figure shows that the three points A, P and B are collinear.

Shifting the line AB at the origin as shown in the figure; points O, P, Q are collinear. Hence,

$\arg(z-z_{2})=\arg(z_{1}-z_{2})$ or

$\arg {\frac{z-z_{2}}{z_{1}-z_{2}}}=0$

$\Longrightarrow \frac{z-z_{2}}{z_{1}-z_{2}}$ is purely real.

$\frac{z-z_{2}}{z_{1}-z_{2}}=\frac{\overline{z-z_{2}}}{z_{1}-z_{2}}$

or, $\frac{z-z_{2}}{z_{1}-z_{2}}=\frac{\overline{z}-\overline{z_{2}}}{\overline{z_{1}}-\overline{z_{2}}}$ call this as Equation 1.

$\left | \begin{array}{ccc} z & \overline{z} & 1 \\ z_{1} & \overline{z_{1}} & 1\\ z_{2} & \overline{z_{2}} & 1 \end{array} \right |=0$ . Call this as Equation 2.

Hence, from (2), if points $z_{1}$ , $z_{2}$ , $z_{3}$ are collinear, then

$\left | \begin{array}{ccc} z_{1} & \overline{z_{1}} & 1 \\ z_{2} & \overline{z_{2}} & 1 \\ z_{3} & \overline{z_{3}} & 1 \end{array} \right |=0$ .

Equation (2) can also be written as

$(\overline{z_{1}}- \overline{z_{2}}) - (z_{1}-z_{2})\overline{z}+z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}=0$

$\Longrightarrow i(\overline{z_{1}}-\overline{z_{2}})z-(z_{1}-z_{2})\overline{z} + z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}}=0$

$\Longrightarrow \overline{a}z + a\overline{z}+b=0$ let us call this Equation 3.

where $a=-i(z_{1}-z_{2})$ and $b=i(z_{1}\overline{z_{2}}-z_{2}\overline{z_{1}})=i 2i \times \Im (z_{1} \overline{z_{2}})$ , which in turn equals

$-2 \times \Im(z_{1}\overline{z_{2}})$ , which is a real number.

Slope of the given line

In Equation (3), replacing z by $x+iy$ , we get $(x+iy)\overline{a} + (x-iy)a+b=0$ ,

$\Longrightarrow (a+\overline{a})x + iy(\overline{a}-a)+b=0$

Hence, the slope $= \frac{a+\overline{a}}{i(a-\overline{a})}=\frac{2 \Re(a)}{2i \times \Im(a)}=-\frac{\Re(a)}{\Im(a)}$

Equation of a line parallel to the line $z \overline{a}+\overline{z}a+b=0$ is $z \overline{a} + \overline{z} a + \lambda=0$ (where $\lambda$ is a real number).

Equation of a line perpendicular to the line $z\overline{a}+\overline{z}a+b=0$ is $z\overline{a}+\overline{z} a + i \lambda=0$ (where $\lambda$ is a real number).

Equation of a perpendicular bisector

Consider a line segment joining $A(z_{1})$ and $B(z_{2})$ . Let the line L be its perpendicular bisector. If $P(z)$ be any point on L, then we have (see attached fig 2)

$PA=PB \Longrightarrow |z-z_{1}|=|z-z_{2}|$

or $|z-z_{1}|^{2}=|z-z_{2}|^{2}$

or $(z-z_{1})(\overline{z}-\overline{z_{1}})=(z-z_{2})(\overline{z}-\overline{z_{2}})$

Here, $a= z_{2}-z_{1}$ and $b=z_{1}\overline{z_{1}}-z_{2} \overline{z_{2}}$

Distance of a given point from a given line:

(See attached Fig 3).

Let the given line be $z \overline{a} + \overline{z} a + b=0$ and the given point be $z_{c}$ . Then,

$z_{c}=x_{c}+iy_{c}$

Replacing z by $x+iy$ in the given equation, we get

$x(a+\overline{a})+iy(\overline{a}-a)+b=0$

Distance of $(x_{c},y_{c})$ from this line is

$\frac{|x_{c}(a+\overline{a})+iy_{c}(\overline{a}-a)+b|}{\sqrt{(a+\overline{a})^{2}-(a-\overline{a})^{2}}}$

which in turn equals

$\frac{z_{c}\overline{a}+\overline{z_{c}}a+b}{\sqrt{4(\Re(a))^{2}+4(\Im(a))^{2}}}$ which is equal to finally

$\frac{|z_{c}\overline{a}+\overline{z_{c}}a+b|}{2|a|}$ .

More later,

Nalin Pithwa

By nkpithwa | Also posted in Complex Numbers | Tagged algebra for IITJEE, complex numbers for IITJEE, Tewani | Comments (0)

A Little Note on Complex Numbers and Geometry for IITJEE Maths

February 27, 2016 – 6:13 am

Note:

In acute triangle, orthocentre (H), centroid (G), and circumcentre (O) are collinear and $HG:GO = 2:1$
Centroid of the triangle formed by points $A(z_{1})$ , $B(z_{2})$ , $C(z_{3})$ is $(z_{1}+z_{2}+z_{3})/3$ .
If the circumcentre of a triangle formed by $z_{1}$ , $z_{2}$ and $z_{3}$ is origin, then its orthocentre is $z_{1}+z_{2}+z_{3}$ (using 1).

Example 1:

Find the relation if $z_{1}$ , $z_{2}$ , $z_{3}$ , $z_{4}$ are the points of the vertices of a parallelogram taken in order.

Solution:

As the diagonals of a parallelogram bisect each other, the affix of the mid-point of AC is same as the affix of the mid-point of BD. That is,

$\frac{z_{1}+z_{3}}{2}=\frac{z_{2}+z_{3}}{2}$

or $z_{1}+z_{3}=z_{2}+z_{4}$

Example 2:

if $z_{1}$ , $z_{2}$ , $z_{3}$ are three non-zero complex numbers such that $z_{3}=(1-\lambda)z_{1}+\lambda z_{2}$ where $\lambda \in \Re - \{ 0 \}$ then prove that the points corresponding to $z_{1}$ , $z_{2}$ , and $z_{3}$ are collinear.

Solution:

$z_{3}=(1-\lambda)z_{1}+z_{2} =$ latex \frac{(1-\lambda)z_{1}+\lambda z_{2}}{1-\lambda +\lambda}$.

Hence, $z_{3}$ divides the line joining $A(z_{1})$ and $B(z_{2})$ in the ratio $\lambda : (1-\lambda)$ . Thus, the given points are collinear.

Homework:

Let $z_{1}$ , $z_{2}$ , $z_{3}$ be three complex numbers and a, b, c be real numbers not all zero, such that $a+b+c=0$ and $az_{1}+bz_{2}+cz_{3}=0$ . Show that $z_{1}$ , $z_{2}$ , $z_{3}$ are collinear.
In triangle PQR, $P(z_{1})$ , $Q(z_{2})$ , and $R(z_{3})$ are inscribed in the circle $|z|=5$ . If $H(z^{*})$ be the orthocentre of triangle PQR, then find $z^{*}$ .