Question

Let $P(a\sec \theta ,b\tan \theta )$ and $Q(a\sec \varphi ,b\tan \varphi )$, where $\theta +\varphi =\frac{\pi }{2}$, 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 the normals at $P$ and $Q$, then $k$ is equal to

• A. $\frac{a^2+b^2}{a}$
• B. $-\left(\frac{a^2+b^2}{a}\right)$
• C. $\frac{a^2+b^2}{b}$
• D. $-\left(\frac{a^2+b^2}{b}\right)$

Answer 1

Answer: (D)

$-\left(\frac{a^2+b^2}{b}\right)$

The equation of a normal to hyperbola at a point $\theta$ is $\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2$
Solving the intersection of two normals at $\theta$ and $\varphi$, we get $by(\cos ec\theta -\cos ec\varphi )=(a^2+b^2)(\sec \theta -\sec \varphi )$
We also have that $\theta +\varphi =\frac{\pi }{2}$
$\Rightarrow by(\sec \theta -\sec \varphi )=(a^2+b^2(\sec \varphi -\sec \theta )$
$\Rightarrow y=\frac{-(a^2+b^2)}{b}$
Hence, option 'D' is correct.