Question

Let P $(asec\theta ,btan\theta )$ and Q $(asec\varphi ,btan\varphi )$, 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 & 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 $-\left(\frac{a^2+b^2}{b}\right)$

Normal at $\theta ,\varphi$ are
$\{\begin{array}{ccc}axcos\theta & +& bycot\theta =a^2+b^2\\ axcos\varphi & +& bycot\varphi =a^2+b^2\end{array}$
where $\varphi =\frac{\pi }{2}-\theta$ and these passes through (h, k).
$\therefore ahcos\theta +bkcot\theta =a^2+b^2$ .....(i)
$ahsin\theta +bktan\theta =a^2+b^2$ .....(ii)
Multiply (i) by $sin\theta$ & (ii) by $cos\theta$ & subtract them,
we get
$\Rightarrow (bk+a^2+b^2)(sin\theta -cos\theta )=0$
$k=-\left(\frac{a^2+b^2}{b}\right)$
Hence, option 'D' is correct.