Question

The locus of the poles of normal chords of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, is:

• A. $\frac{x^2}{a^4}+\frac{y^2}{b^4}=a^2+b^2$
• B. $\frac{x^2}{a^4}+\frac{y^2}{b^4}=a^2-b^2$
• C. $\frac{a^6}{x^2}+\frac{b^6}{y^2}=(a^2-b^2{)}^2$
• D. $\frac{a^4}{x^2}+\frac{b^4}{y^2}=(a^2-b^2{)}^2$

Answer 1

The correct option is C $\frac{a^6}{x^2}+\frac{b^6}{y^2}=(a^2-b^2{)}^2$

Equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(1)
Let $(h,k)$ be the polar
Now polar of $(h,k)$ w.r.t the ellipse is given by $\frac{xh}{a^2}+\frac{yh}{b^2}=1$ ...(2)
It is normal to the ellipse then it must be identical with $ax\sec \theta -by\csc \theta =a^2-b^2$ ...(3)
Hence comparing (2) and (3), we get
$\frac{\left(\frac{h}{a^2}\right)}{a\sec \theta }=\frac{\left(\frac{k}{b^2}\right)}{-b\csc \theta }=\frac{1}{a^2-b^2}$
$\Rightarrow \cos \theta =\frac{a^3}{h(a^2-b^2)}$ and $\sin \theta =\frac{{-b}^3}{k(a^2-b^2)}$
Squaring adding we get
$1=\frac{1}{{(a^2-b^2)}^2}(\frac{a^6}{h^2}+\frac{b^6}{k^2})$
Therefore required locus of $(h,k)$ is
$\frac{a^6}{x^2}+\frac{b^6}{y^2}=(a^2-b^2)$