Question

The locus of poles w.r.t the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ of tangents to the ellipse $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$ is :

• A. $\frac{{\alpha }^2{x}^2}{a^4}+\frac{{\beta }^2{y}^2}{b^4}=1$
• B. $\frac{{\alpha }^2{x}^2}{a^4}-\frac{{\beta }^2{y}^2}{b^4}=1$
• C. $\frac{{\alpha }^2{x}^2}{b^4}-\frac{{\beta }^2{y}^2}{a^4}=1$
• D. $\frac{{\alpha }^2{x}^2}{b^4}+\frac{{\beta }^2{y}^2}{a^4}=1$

Answer 1

The correct option is A $\frac{{\alpha }^2{x}^2}{a^4}+\frac{{\beta }^2{y}^2}{b^4}=1$

Given hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Equation of the Polar of point P $(X,Y)$ wrt to the hyperbola is

$\frac{xX}{a^2}-\frac{yY}{b^2}=1$

The tangent will touch the ellipse if $a^2{l}^2+b^2{m}^2=n^2$

We can see that $l=\frac{X}{a^2},m=-\frac{Y}{b^2},n=-1$

putting values for the ellipse we get,

${\alpha }^2\left(\frac{X^2}{a^4}\right)+{\beta }^2\left(\frac{Y^2}{b^4}\right)={(-1)}^2\Rightarrow {\alpha }^2\left(\frac{x^2}{a^4}\right)+{\beta }^2\left(\frac{y^2}{b^4}\right)=1$