Question

The area of the triangle formed by the lines $x-y=0,x+y=0$ and any tangent to the hyperbola $x^2-y^2=a^2$ is:

• A. $2a^2$
• B. $4a^2$
• C. $a^2$
• D. None of these

Answer 1

Answer: (C)

$a^2$

Any tangent at $P(a\sec \theta ,b\tan \theta )$ to the hyperbola $x^2-y^2=a^2$ is

$x\sec \theta -y\tan \theta =a$ ...(1)

Given lines are $x-y=0$ ...(2)

and $x+y=0$ ...(3)

Solve (1) and (2), (2) and (3), (3) and (1), we get vertices of the triangle as

$(\frac{a}{\sec \theta -\tan \theta },\frac{a}{\sec \theta -\tan \theta }),(\frac{a}{\sec \theta +\tan \theta },\frac{-a}{\sec \theta +\tan \theta })$ and $(0,0)$

$\therefore$ Area of the triangle $=\frac{1}{2}|x_1{y}_2-x_2{y}_1|$

$=\frac{a^2}{2}[\frac{-1}{{\sec }^2\theta -{\tan }^2\theta },\frac{1}{{\sec }^2\theta -{\tan }^2\theta }]=\frac{a^2}{2}(-2)=a^2$ in magnitudes