Question

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

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

Answer 1

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

$Gg=2PC$

Let $P(x_1,y_1)$ on $x^2-y^2=4$.
Normal at $P$ is $x_{1}y+x{y}_1=2x_1{y}_1$
let $G\equiv (2x_1,0)$ and $g\equiv (0,2y_1)$
$PG=\sqrt{(2x_1-x_1{)}^2+y_1^2}=\sqrt{x_1^2+y_1^2}=PC$
$Pg=\sqrt{x_1^2+(2y_1-y_1{)}^2}=\sqrt{x_1^2+y_1^2}=PC$
and clearly $PG=Pg=PC$
$Gg=\sqrt{(2x_1{)}^2+(2y_1{)}^2}=2\sqrt{x_1^2+y_1^2}=2PC$