Question

Let $P(asec\theta ,b\tan \theta )$ and $Q(asec\varphi ,b\tan \varphi ),$ where $\theta +\varphi =\frac{\pi }{2},$ be two points on the hyperbola $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{b^2}\right)=1$. If $(h,k)$ is the point of intersection of the normal at $P$ and $Q$, then$k=$

• A. $\frac{[a^2+b^2]}{a}$
• B. $–\frac{[a^2+b^2]}{a}$
• C. $\frac{[a^2+b^2]}{b}$
• D. $–\frac{[a^2+b^2]}{b}$

Answer 1

The correct option is D

$–\frac{[a^2+b^2]}{b}$

Explanation for the correct option:

Finding the value of k:

The slope of normal to $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{b^2}\right)=1$at $(asec\theta ,b\tan \theta )$

$$\begin{gathered} \frac{2x}{a^2}-\frac{2y}{b^2}\times \frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\left(\frac{b^2}{a^2}\right)\left(\frac{x}{y}\right) \end{gathered}$$

Slope for normal at the point $(asec\theta ,b\tan \theta )$ will be $\frac{–[a^{2}b\tan \theta ]}{b^{2}asec\theta }=\frac{–a}{b\sin \theta }$

∴ Equation of normal at$(asec\theta ,b\tan \theta )$ is $y–b\tan \theta =–\left(\frac{a}{b}\right)\sin \theta (x–asec\theta )$

$$\begin{gathered} \Rightarrow (a\sin \theta )x+by=(a^2+b^2)\tan \theta \\ \Rightarrow ax+b\cos ec\theta =(a^2+b^2)sec\theta \dots ..\left(i\right) \end{gathered}$$

Similarly, the equation of normal to $\left(\frac{x^2}{a^2}\right)-\left(\frac{y^2}{b^2}\right)=1$at $(asec\varphi ,b\tan \varphi ),$ $y=(a^2+b^2)(sec\theta –sec\varphi )\dots .\left(ii\right)$

On subtracting equation $\left(ii\right)$ from equation$\left(i\right)$, $b(\cos ec\theta –\cos ec\varphi )y=(a^2+b^2)(sec\theta –sec\varphi )$

$$\begin{gathered} y=\frac{(a^2+b^2)}{b}.\left[\frac{\sec \theta –\sec \varphi }{\cos ec\theta –\cos ec\varphi }\right] \\ =\frac{(a^2+b^2)}{b}.\left[\frac{\sec \theta –\sec \left(\frac{\pi }{2}-\theta \right)}{\csc \theta –\csc \left(\frac{\pi }{2}-\theta \right)}\right]\left[\because \varphi +\theta =\frac{\pi }{2}\right] \\ =\frac{(a^2+b^2)}{b}.\left(–1\right) \\ \therefore y=–\frac{[a^2+b^2]}{b} \end{gathered}$$

Therefore, the correct answer is option (D).