Question

A normal to the hyperbola, $4x^2-9y^2=36$ meets the co-ordinate axes $x$ and $y$ at $A$ and $B$, respectively. If the parallelogram $OAPB$ ($O$ being the origin) is formed, then the locus of $P$ is :

• A. $4x^2-9y^2=121$
• B. $4x^2+9y^2=121$
• C. $9x^2-4y^2=169$
• D. $9x^2+4y^2=169$

Answer 1

The correct option is C $9x^2-4y^2=169$

The equation of hyperbola is $4x^2-9y^2=36$
$\Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=1$
where $a=3$ and $b=2$
Equation of normal at $(3\sec \theta ,2\tan \theta )$ is
$\frac{x}{\frac{13\sec \theta }{3}}+\frac{y}{\frac{13\tan \theta }{2}}=1$
So, co-ordinates of $A$ and $B$ are $A(\frac{13}{3}\sec \theta ,0)$ and $B(0,\frac{13}{2}\tan \theta )$
In the parallelogram $OAPB$, as the adjacent angles are ${90}^{\circ },OAPB$ is going to be a rectangle.


$\Rightarrow h=\frac{13\sec \theta }{3};k=\frac{13\tan \theta }{2}$(since opposite side lengths are equal)
Applying ${\sec }^2\theta -{\tan }^2\theta =1$, we get
$9h^2-4k^2=169$
Hence, locus is $9x^2-4y^2=169$