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 $OABP$ ($O$ 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 $4x^2-9y^2=36$
where $a=3$ and $b=2$
Let the tangent pass through point $P(a\sec \theta ,b\tan \theta )$
So, point $P$ is $(3\sec \theta ,2\tan \theta )$
Equation of tangent on hyperbola is
$4x\left(\right(3\sec \theta )-9y(2\tan \theta )=36$
$\Rightarrow 2\sec \theta x-3\tan \theta y=6$
slope$=\frac{2}{3\sin \theta }$
slope of normal$=-\frac{3\sin \theta }{2}$
Equation of normal is
$y-2\tan \theta =-\frac{3\sin \theta }{2}(x-3\sec \theta )$
$\Rightarrow \frac{x}{\frac{13\sec \theta }{3}}-\frac{y}{\frac{13\tan \theta }{2}}=1$
So, co-orinates of $A$ and $B$ are $A(\frac{13}{3}\sec \theta ,0)$ and $B(0,\frac{13}{2}\tan \theta )$

From the midpoint of diagonals of parallelogram $OABP$
mid point of $AP\equiv$ mid point of $OB$
$\Rightarrow M(\frac{\frac{13}{3}\sec \theta +h}{2},\frac{k}{2})\equiv M(0,\frac{\frac{13}{2}\tan \theta }{2})$
$\therefore \frac{\frac{13}{3}\sec \theta +h}{2}=0,\frac{\frac{13}{2}\tan \theta }{2}=\frac{k}{2}$
$\Rightarrow \sec \theta =-\frac{3h}{13},\tan \theta =\frac{2k}{13}$
$\therefore {\sec }^{\theta }-{\tan }^2\theta =1$
$\Rightarrow \frac{9h^2}{169}-\frac{4k^2}{169}=1$
$\Rightarrow 9h^2-4k^2=169$
Hence, $9x^2-4y^2=169$