Question

Find the equation of the locus of the poles of the normal chords of the ellipse $x^2/a^2+y^2/b^2=1$.

• A. $x^2{y}^2(a^2-b^2{)}^2=a^4{y}^2+b^4{x}^2$
• B. $x^2{y}^2(a^2-b^2{)}^2=a^6{y}^2+b^4{x}^2$
• C. $x^2{y}^2(a^2-b^2{)}^2=a^6{y}^2+b^6{x}^2$
• D. None of the above

Answer 1

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

Equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Let $(h,k)$ be the polar

Now polar of $(h,k)$ wrt the ellipse is given by $\frac{xh}{a^2}+\frac{yk}{b^2}=1$

It is normal to the ellipse then it must be identical with $axsec\theta -bycosec\theta =a^2-b^2$

Comparing both the equations we get

$\frac{h/a^2}{asec\theta }=\frac{k/b^2}{-bcosec\theta }=\frac{1}{a^2-b^2}$

$\cos \theta =\frac{a^3}{h(a^2-b^2)}$ and $\sin \theta =\frac{-b^3}{k(a^2-b^2)}$

Squaring and adding we get

$1=\frac{1}{(a^2-b^2{)}^2}\times \frac{a^6}{h^2}+\frac{b^6}{k^2}$

Therefore required locus of $(h,k)$ is

$\frac{a^6}{x^2}+\frac{b^6}{y^2}=(a^2-b^2)$

$x^2{y}^2(a^2-b^2{)}^2=a^6{y}^2+b^6{x}^2$

So option C is the correct answer