Question

The locus of poles with respect to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ of any tangent to the auxilary circle is the curve $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{c}$, then $c$ is:

• A. $a^2$
• B. $b^2$
• C. $a^4$
• D. $b^4$

Answer 1

The correct option is A $a^2$

Equation of the auxilliary circle is
$x^2+y^2=a^2$
Let (h,k) be the pole , then the equation of the polar
of (h,k) with respect to the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{hx}{a^2}+\frac{ky}{b^2}=1$
Since the above equation is tangent to the circle
$\frac{-1}{\sqrt{\frac{h^2}{a^2}+\frac{k^2}{b^2}}}=\pm a$
$\frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}$
Locus of (h,k) is
$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}$
Therefore value of $c=a^2$
Option A