Question

If in a rectangular hyperbola normal at any point $P$ meets the axes in $G$ and $g$ and $c$ be the center of hyperbola, then

• A. $PG=Pg$
• B. $PG=Pc$
• C. $Pg=Pc$
• D. None of these

Answer 1

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

The correct options are
A $PG=Pg$
B $PG=Pc$
C $Pg=Pc$

Let the equation to the rectangular hyperbola be, $x^2-y^2=a^2$
If $P$ be any point on it as $(a\sec \varphi ,a\tan \varphi )$,
then the equation to the normal at $P$ will be given by $x\sin \varphi +y=2a\tan \varphi$
Its intersection with $y=0$ gives $x=2a\sec \varphi$.
As the point of intersection is $G$ and $C$, the center of the hyperbola hence.
$CG=2a\sec \varphi$
Again the intersection of the normal with $x=0$ gives $y=2a\tan \varphi$
If $g$ denote this point of intersection, we have $Cg=2a\tan \varphi$
Now ${PG}^2={(a\sec \varphi -2a\sec \varphi )}^2+{(a\tan \varphi -0)}^2=a^2({\sec }^2\varphi +{\tan }^2\varphi )$
${Pg}^2={(a\sec \varphi -0)}^2+{(a\tan \varphi -2a\tan \varphi )}^2=a^2({\sec }^2\varphi +{\tan }^2\varphi ){PC}^2={(a\sec \varphi -0)}^2+{(a\tan \varphi -0)}^2=a^2({\sec }^2\varphi +{\tan }^2\varphi )$
Hence $PG=Pg=PC.$