Question

$P$ is a point on the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ and $Q$ is its corresponding point on the auxiliary circle. lf the locus of the point of intersection of the normal at $P$ and $Q$ to the respective curves is circle, then its radius is:

• A. $5$
• B. $7$
• C. $6$
• D. $8$

Answer 1

The correct option is B $7$

$y=x\tan \theta$(Normal to circle )
$4x\sec \theta -3y\text{cosec}\theta =7$
(normal to ellipse)
$4x\sec \theta -3x\sec \theta =7$
$x\sec \theta =7$
$x=7\cos \theta$
$y=7\sin \theta$
$\therefore {\left(\frac{x}{7}\right)}^2+{\left(\frac{y}{7}\right)}^2=1$
$\therefore$ Radius of the circle $=7$.