Question

The tangent at any point

P

of a circle

x^{2} + y^{2} = a^{2}

meets the tangent at a fixed point

A (a, 0)

in

T

and

T

is joined to

B

, the other end of the diameter through

A

, Prove that the locus of intersection of

A P

and

B T

is an ellipse whose eccentricity is

\frac{1}{\sqrt{2}}

.

Accepted Answer

The equation of the circle be,

$x^{2} + y^{2} = a^{2}$

Its radius is $a$ and its centre is the origin $(0, 0)$ .

Let $A$ be a point $P$ on the circle and $O A$ be the $x$ -axis. Thus, we get,

$\Rightarrow A$ $(a, 0)$

$\Rightarrow B$ $(- a, 0)$

The equation of the tangent at $P$ can be given as,

$x cosθ + y sinθ = a$

And, the equation of the tangent at $A$ can be given as,

$x = a$

Solving both the equations, we get,

$y = \frac{a (1 - cosθ)}{sinθ}$

$y = \frac{2 a \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}$

$y = a \tan \frac{θ}{2}$

Hence, the coordinates of $T$ are $(a, a \tan \frac{θ}{2})$ .

Now, the equation of the line $A P$ is given as,

$y - 0 = \frac{a \sin θ - 0}{a \cos θ - a} (x - a)$

$y - 0 = - (x - a) \cot \frac{θ}{2}$

$\cot \frac{θ}{2} = - \frac{y}{x - a} . . . (1)$

Similarly, the equation of $B T$ is given as,

$y - 0 = \frac{a \tan \frac{θ}{2} - 0}{a - (- a)} (x + a)$

$y - 0 = \frac{1}{2} (x + a) \tan \frac{θ}{2}$

$\tan \frac{θ}{2} = \frac{2 y}{x + a} . . . (2)$

To find the locus of the point of intersection of $A P$ and $B T$ , we need to eliminate the variable $θ$ . So, by multiplying equations $(1)$ and $(2)$ , we get,

$\Rightarrow \frac{- 2 y^{2}}{x^{2} - a^{2}} = 1$

$\Rightarrow 2 y^{2} + x^{2} = a^{2}$

$\Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{\frac{a^{2}}{2}} = 1$ (which is the equation of an ellipse)

$\Rightarrow e = \sqrt{(\frac{A^{2} - B^{2}}{A^{2}})}$

$e = \sqrt{(\frac{a^{2} - \frac{a^{2}}{2}}{a^{2}})}$

$e = \frac{1}{\sqrt{2}}$

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