Question

If the polar with respect to $y^2=4x$ touches the ellipse $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$, the locus of its pole is

• A. $\frac{x^2}{{\alpha }^2}-\frac{y^2}{\left(4{\alpha }^2/{\beta }^2\right)}=1$
• B. $\frac{x^2}{{\alpha }^2}+\frac{{\beta }^2{y}^2}{4a^2}=1$
• 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 be $P(h,k)$. thus it's polar w.r.t $y^2=4x$ is given by, $T=0\Rightarrow ky=2(x+h)$
or $y=\frac{2}{k}x+\frac{2h}{k}$.

This line touch ellipse, $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$
Thus applying condition of tangency, $c^2=a^2{m}^2+b^2$
$\Rightarrow \frac{4h^2}{k^2}=\frac{4{\alpha }^2}{k^2}+{\beta }^2\Rightarrow \frac{h^2}{{\alpha }^2}-\frac{k^2}{\left(4{\alpha }^2/{\beta }^2\right)}=1$
Hence required locus of $P(h,k)$ is $\frac{x^2}{{\alpha }^2}-\frac{y^2}{\left(\frac{4{\alpha }^2}{{\beta }^2}\right)}=1$