Question

A chord $MP$ parallel to the latus rectum of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ with centre at $O(0,0)$ intersects the auxiliary circle at $Q$. Then the locus of the point of intersection of normals at $P$ and $Q$ to the respective curve is

• A. $x^2+y^2=8$
• B. $x^2+y^2=34$
• C. $x^2+y^2=64$
• D. $x^2+y^2=15$

Answer 1

The correct option is C $x^2+y^2=64$


$\frac{x^2}{25}+\frac{y^2}{9}=1;x^2+y^2=25$
$P(5\cos \theta ,3\sin \theta ),Q(5\cos \theta ,5\sin \theta )$

Equation of normal at $Q$ is
$y-0=\tan \theta (x-0)$
$\Rightarrow y=x\tan \theta$

Equation of normal at $P$ is
$\frac{5x}{\cos \theta }-\frac{3y}{\sin \theta }=16$

Solving the two equations, we get
$y=8\sin \theta ,x=8\cos \theta$
$x^2+y^2=64$ which represents a circle.