Question

If the polar with respect to $y^2=4x$ touches the ellipse $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}$ , the locus of its pole is

• A. $\frac{x^2}{{\alpha }^2}-\frac{y^2}{\left(4{\alpha }^2/{\beta }^2\right)}=1$
• B. $\frac{x^2}{{\alpha }^2}-\frac{y^2}{\left(4\right)}={\beta }^2$
• C. ${\alpha }^2{x}^2+{\beta }^2{y}^2=1$
• D. None of these

Answer 1

The correct option is A $\frac{x^2}{{\alpha }^2}-\frac{y^2}{\left(4{\alpha }^2/{\beta }^2\right)}=1$

Let pole is $(h,k)$

It's polar with respect to $y^2=4x$ is given by $T=0$

$ky=2(x+h)$

or $y=\frac{2x}{k}+\frac{2h}{k}$

This line touches the ellipse $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$

On applying the condition of tangency we get

$c^2=a^2{m}^2+b^2$

$\frac{4h^2}{k^2}=\frac{4{\alpha }^2}{k^2}+{\beta }^2$

$\frac{h^2}{{\alpha }^2}-\frac{k^2}{\left(4{\alpha }^2\right)/\left({\beta }^2\right)}=1$

So the required locus is

$\frac{x^2}{{\alpha }^2}-\frac{y^2}{\left(4{\alpha }^2\right)/\left({\beta }^2\right)}=1$

So option A is the correct answer