Question

Any ordinate $MP$ of an ellipse meets the auxiliary circle in $Q$. The locus of the point of intersection of the normals at $P$ and $Q$ is:

• A. $x^2+y^2=a^2+b^2$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=a^2+b^2$
• C. $x^2+y^2={(a+b)}^2$
• D. $x^2+y^2=2(a^2+b^2)$

Answer 1

The correct option is C $x^2+y^2={(a+b)}^2$

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

It passes through point $(h,k)$, then

$\frac{ah}{\cos \theta }-\frac{bk}{\sin \theta }=a^2-b^2$ ----------(1)

Normal at Q passes through the centre of the ellipse

$\therefore$ Normal at Q: $\frac{y}{x}=tan\theta$

It also passes through $(h,k)$: $\frac{k}{h}=tan\theta$

$\therefore \cos \theta =\frac{h}{\sqrt{h^2+k^2}}and\sin \theta =\frac{k}{\sqrt{h^2+k^2}}$

Putting the values in equation (1) we get:

$a\sqrt{h^2+k^2}-b\sqrt{h^2+k^2}=a^2-b^2$

Replacing $(h,k)$ with $(x,y)$

$\therefore a\sqrt{x^2+y^2}-b\sqrt{x^2+y^2}=a^2-b^2$

$\Rightarrow \sqrt{x^2+y^2}(a-b)=a^2-b^2$

$\Rightarrow \sqrt{x^2+y^2}=a+b$

Squaring both sides we get: $\therefore x^2+y^2={(a+b)}^2$