Question

Let $P$ be a point on the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. An ordinate $MP$ of the ellipse meets 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. $\frac{x^2}{16}+\frac{y^2}{36}=1$
• B. $\frac{x^2}{36}+\frac{y^2}{16}=1$
• C. $x^2+y^2=64$
• D. $x^2+y^2=16$

Answer 1

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

$P(a\cos \theta ,b\sin \theta )$

$Q(a\cos \theta ,a\sin \theta )$

Normal at $P=\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$

$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$

Normal at $Q=\frac{y}{x}=\frac{y_1}{x_1}=\tan \theta$

$y=x\tan \theta$

Point of intersections

$\frac{ax}{\cos \theta }-\frac{bx\tan \theta }{\sin \theta }=a^2-b^2$

$x[\frac{a}{\cos \theta }-\frac{b}{\cos \theta }]=a^2-b^2\Rightarrow x=(a+b)\cos \theta$

$y=(a+b)\sin \theta$

$x^2+y^2={(a+b)}^2$

$=64$