Question

If the normal at $P$ to the rectangular hyperbola $x^2-y^2=4$ meeets the axes in $G$ and $g$ and $C$ is the centre of the hyperbola, then

• A. $PG=PC$
• B. $Pg=PC$
• C. $PG=Pg$
• D. $Gg=2PC$

Answer 1

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

The correct options are
A $PG=PC$
B $Pg=PC$
C $PG=Pg$
D $Gg=2PC$

Normal at point $P(2\sec \theta ,2\tan \theta )$ is
$\frac{2x}{\sec \theta }+\frac{2y}{\tan \theta }=8$
It meets the axes at points $G(4\sec \theta ,0)$ and $g(0,4\tan \theta )$.
Then,
$PG=\sqrt{4{\sec }^2\theta +4{\tan }^2\theta }$
$Pg=\sqrt{4{\sec }^2\theta +4{\tan }^2\theta }$
$PC=\sqrt{4{\sec }^2\theta +4{\tan }^2\theta }$
$Gg=\sqrt{16{\sec }^2\theta +16{\tan }^2\theta }$
$=2\sqrt{4{\sec }^2\theta +{\tan }^2\theta }=2PC$