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 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

The correct option is D $-\left(\frac{a^2+b^2}{b}\right)$

Given $P(a\sec \theta ,b\tan \theta )$ and $Q(a\sec \varphi ,b\tan \varphi ).$
The equation of normal at point $P$ is
$\frac{a^{2}x}{a\sec \theta }+\frac{b^{2}y}{b\tan \theta }=a^2+b^2$
$\Rightarrow ax\sin \theta +by=(a^2+b^2)\tan \theta \cdots \left(1\right)$
Similarly the equation of normal at point $Q$ is
$ax\sin \varphi +by=(a^2+b^2)\tan \varphi \cdots \left(2\right)$

On mulitplying $\left(1\right)$ by $\sin \varphi$ and $\left(2\right)$ by $\sin \theta$ , we get
$ax\sin \theta \sin \varphi +by\sin \varphi =(a^2+b^2)\tan \theta \sin \varphi$
$ax\sin \varphi \sin \theta +by\sin \theta =(a^2+b^2)\tan \varphi \sin \theta$

On subtracting eqn $\left(2\right)$ from $\left(1\right)$, we get
$by(\sin \varphi -\sin \theta )=(a^2+b^2)(\tan \theta \sin \varphi -\tan \varphi \sin \theta )$
$\therefore y=k=\frac{a^2+b^2}{b}\cdot \frac{\tan \theta \sin \varphi -\tan \varphi \sin \theta }{\sin \varphi -\sin \theta }$
$\because \theta +\varphi =\frac{\pi }{2}\Rightarrow \varphi =\frac{\pi }{2}-\theta$
$\Rightarrow \sin \varphi =\cos \theta \text{and}\tan \varphi =\cot \theta$
$\therefore y=k=\frac{a^2+b^2}{b}\cdot \frac{\tan \theta \cos \theta -\cot \theta \sin \theta }{\cos \theta -\sin \theta }$
$=\frac{a^2+b^2}{b}\left(\frac{\sin \theta -\cos \theta }{\cos \theta -\sin \theta }\right)=-\frac{(a^2+b^2)}{b}$