Question

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

• A. $\frac{x^2}{{\alpha }^2}-\frac{y^2}{4{\alpha }^2/{\beta }^2}=1$,
• B. $\frac{x^2}{{\alpha }^2}-\frac{y^2\beta }{4{\alpha }^2}=1$,
• C. $\frac{x^2}{{\alpha }^2}-\frac{y^2}{{\alpha }^2/{\beta }^2}=1$,
• D. None of the above

Answer 1

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

Let pole be P(h,k). Thus its polar wrt to $y^2=4x$ is given by $T=0$

$ky=2(x+h)$

or

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

Given this line touches the ellipse

$\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$

Thus applying the condition of tangency we get

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

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

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

Hence the required locus will be

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