Question

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

• A. $\frac{x^{2}}{\alpha^{2}} - \frac{y^{2}}{(4a^{2}\alpha^{2}/\beta^{2})} = 1$
• 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

Answer: (A)

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

Answer 2

Let $P(h,k)$ be the pole. Then the equation of the polar is $ky = 2a(x + h)$ or $y = \frac{2a}{k}x + \frac{2ah}{k}$.

This touches $\frac{x^{2}}{\alpha^{2}} + \frac{y^{2}}{\beta^{2}} = 1$, So $\left( \frac{2ah}{k} \right)^{2} = \alpha^{2}\left( \frac{2a}{k} \right)^{2} + \beta^{2},$

(using $c^{2} = a^{2}m^{2} + b^{2}$)

⇒$4a^{2}h^{2} = 4a^{2}\alpha^{2} + k^{2}\beta^{2}$. So, locus of $(h,k)$ is

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