Question

$PN$ is the ordinate of any point $P$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $A{A}^{\prime }$ is its transverse axis. If $Q$ divides $AP$ in the ratio $a^2:b^2$, then $NQ$ is :

• A. $\perp$ to $A^{\prime }P$
• B. parallel to $A^{\prime }P$
• C. $\perp$ to $OP$
• D. None of these

Answer 1

The correct option is A $\perp$ to $A^{\prime }P$

Let $P\equiv (a\sec \theta ,b\tan \theta )$
Then, $N\equiv (a\sec \theta ,0)$
Since $Q$ divides $AP$ in the ratio $a^2:b^2$
$\therefore$ coordinates of $Q$ are $=(\frac{a{b}^2+a^2\sec \theta }{a^2+b^2},\frac{a^{2}b\tan \theta }{a^2+b^2})$
Slope of $A^{\prime }P=\frac{b\tan \theta }{a(\sec \theta +1)}$
Slope of $QN=\frac{a^{2}b\tan \theta }{a{b}^2+a^3\sec \theta -a^3\sec \theta -a{b}^2\sec \theta }=\frac{a^{2}b\tan \theta }{a{b}^2(1-\sec \theta )}$
$\therefore$ Slope of $A^{\prime }P\times$ slope of $QN=\frac{a^2{b}^2{\tan }^2\theta }{-a^2{b}^2{\tan }^2\theta }=-1$
$\therefore QN$ is $\perp$ to $A^{\prime }P$