Question

The tangent at any point of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ cuts of a triangle from the asymptotes and that the portion of it intercepted between the asymptotes is bisected at the point of contact, then area of this triangle is given by:

• A. $4ab$ sq.units
• B. $2ab$ sq.units
• C. $a^2{b}^2$ sq.units
• D. $ab$ sq.units

Answer 1

Answer: (D)

$ab$ sq.units

The combined equation of the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(\ast )$ is given by
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$
$y=\pm \frac{b}{a}x$are equation of asymptotes... (* *)
let $P(a\sec \theta ,b\tan \theta )$ be any point on the hyperbola (*)
$\therefore$ equation of tangent of $P$ is
$\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1$ ....(* * *)
Solving (* * ) & (* * *) for the point of intersection of the asymptotes and the tangent at $P$
$\therefore A[a(\sec \theta +\tan \theta ),b(\sec \theta +\tan \theta )],B[a(\sec \theta -\tan \theta ),b(\sec \theta -\tan \theta )]$
Area of $△ABC$ is equal
$=\frac{1}{2}\left|\begin{array}{ccc}a(\sec \theta +\tan \theta )& b(\sec \theta +\tan \theta )& 1\\ a(\sec \theta -\tan \theta )& b(\sec \theta -\tan \theta )& 1\\ 0& 0& 1\end{array}\right|$
$=\frac{1}{2}ab\{({\sec }^2\theta -{\tan }^2\theta )\times 2\}$
$=ab\left(1\right)=ab$ sq.units
Hence, option 'D' is correct.