Question

The line

y = m x + a \sqrt{(m^{2} + 1 - e^{2})}; (e^{2} < 1)

, cuts the lines

x = \pm a

in

T

and

T^{'}

. If

S (a e, 0)

and

S^{'} (- a e, 0)

are two other points, show that all these four points are con-cyclic.

Accepted Answer

We have,

$y = m x + a \sqrt{(m^{2} + 1 - e^{2})}$

Putting $x = \pm a$ , then we get

$y = a (\sqrt{(m^{2} + 1 - e^{2})} \pm m)$

Now, we have $S (a e, 0), S^{'} (- a e, 0), T (a, a (\sqrt{(m^{2} + 1 - e^{2})} + m)), T^{'} (- a, a (\sqrt{(m^{2} + 1 - e^{2})} - m))$ .

First we will find the equation of circumcircle of $∆ S S^{'} T$ .

Let the circle be

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

It is passing through $S, S^{'}$ , so

$a^{2} e^{2} + 2 a e g + c = 0$

$a^{2} e^{2} - 2 a e g + c = 0$

Subtracting both, we get

$4 a e g = 0 \Rightarrow g = 0$

Now,

$c = - a^{2} e^{2}$

So,

$x^{2} + y^{2} + 2 f y - a^{2} e^{2} = 0$

It is passing through $T (a, a (\sqrt{(m^{2} + 1 - e^{2})} + m))$ , so

$a^{2} + a^{2} {(\sqrt{(m^{2} + 1 - e^{2})} + m)}^{2} + 2 a (\sqrt{(m^{2} + 1 - e^{2})} + m) f - a^{2} e^{2} = 0$

$\Rightarrow a^{2} (1 - e^{2}) + a^{2} (m^{2} + 1 - e^{2} + m^{2} + 2 m \sqrt{(m^{2} + 1 - e^{2})}) + 2 a (\sqrt{(m^{2} + 1 - e^{2})} + m) f = 0$

$\Rightarrow a^{2} (1 - e^{2}) + a^{2} (2 m^{2} + 1) - a^{2} e^{2} + 2 m a^{2} \sqrt{(m^{2} + 1 - e^{2})} + 2 a (\sqrt{(m^{2} + 1 - e^{2})} + m) f = 0$

$\Rightarrow a^{2} (1 + m^{2} - e^{2} + m \sqrt{(m^{2} + 1 - e^{2})}) + a (\sqrt{(m^{2} + 1 - e^{2})} + m) f = 0$

$\Rightarrow f = \frac{- a ({(\sqrt{1 + m^{2} - e^{2}})}^{2} + m \sqrt{(m^{2} + 1 - e^{2})})}{\sqrt{(m^{2} + 1 - e^{2})} + m}$

$\Rightarrow f = \frac{- a \sqrt{(m^{2} + 1 - e^{2})} ((\sqrt{1 + m^{2} - e^{2}}) + m)}{\sqrt{(m^{2} + 1 - e^{2})} + m}$

$\Rightarrow f = - a \sqrt{(m^{2} + 1 - e^{2})}$

Hence, equation of circle is

$x^{2} + y^{2} - 2 a \sqrt{(m^{2} + 1 - e^{2})} y - a^{2} e^{2} = 0$

Now, putting coordinates of $T^{'} (- a, a (\sqrt{(m^{2} + 1 - e^{2})} - m))$ in $LHS$ , we get

$a^{2} + a^{2} {(\sqrt{(m^{2} + 1 - e^{2})} - m)}^{2} - 2 a [\sqrt{(m^{2} + 1 - e^{2})} \times a (\sqrt{(m^{2} + 1 - e^{2})} - m)] - a^{2} e^{2}$

$= a^{2} + a^{2} ((m^{2} + 1 - e^{2}) + m^{2} - 2 m \sqrt{(m^{2} + 1 - e^{2})}) - 2 a^{2} (m^{2} + 1 - e^{2} - m \sqrt{m^{2} + 1 - e^{2}}) - a^{2} e^{2}$

$= a^{2} - a^{2} (m^{2} + 1 - e^{2}) + a^{2} m^{2} - a^{2} e^{2}$

$= a^{2} - a^{2} (m^{2} + 1) + a^{2} m^{2} = 0$

Hence, $T^{'}$ lies on circumcircle of $∆ T S S^{'}$ , points $T, T^{'}, S, S^{'}$ are concyclic.

Dr. SK Goyal Solutions for Chapter: Circle, Exercise 9: EXERCISE ON LEVEL-II

Dr. SK Goyal Mathematics Solutions for Exercise - Dr. SK Goyal Solutions for Chapter: Circle, Exercise 9: EXERCISE ON LEVEL-II

Questions from Dr. SK Goyal Solutions for Chapter: Circle, Exercise 9: EXERCISE ON LEVEL-II with Hints & Solutions

Learn and Prepare for any exam you want !