Question

If the normal at point $P$ to the rectangular hyperbola $x^2-y^2=4$ meets the transverse and conjugate axes at $A$ and $B$ respectively and $C$ is the centre of the hyperbola, then

• A. $PA=PC$
• B. $PA=PB$
• C. $PB=PC$
• D. $AB=2PC$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $PA=PC$
B $PA=PB$
C $PB=PC$
D $AB=2PC$

Let $P(2\sec \theta ,2\tan \theta )$ be the point.
Then equation of normal at 'P' is given by,
$\frac{x}{\sec \theta }+\frac{y}{\tan \theta }=2\times 2=4$
$\therefore A\equiv (4\sec \theta ,0),B\equiv (0,4\tan \theta )$
Also centre of the hyperbola is $C(0,0)$
Thus $PA=\sqrt{4{\sec }^2\theta +4{\tan }^2\theta }=2\sqrt{{\sec }^2\theta +{\tan }^2\theta }$
$PB=\sqrt{4{\sec }^2\theta +4{\tan }^2\theta }=2\sqrt{{\sec }^2\theta +{\tan }^2\theta }$
$PC=\sqrt{4{\sec }^2\theta +4{\tan }^2\theta }=2\sqrt{{\sec }^2\theta +{\tan }^2\theta }$
and $AB=\sqrt{16{\sec }^2\theta +16{\tan }^2\theta }=4\sqrt{{\sec }^2\theta +{\tan }^2\theta }$