Question

If chord of contact of the tangent drawn from the point $(\alpha ,\beta )$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ touches the circle $x^2+y^2=k^2$, then find the locus of the point $(\alpha ,\beta )$.

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

Answer 1

The correct option is B $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{k^2}$

Equation of chord of contact at $(\alpha ,\beta )$ is $\frac{\alpha x}{a^2}+\frac{\beta y}{b^2}=1$
Since, it touches the circle,
Therefore, $r=p$
$k=\left|\frac{1}{\sqrt{\frac{{\alpha }^2}{a^4}+\frac{{\beta }^2}{b^4}}}\right|$
$\Rightarrow \frac{{\alpha }^2}{a^4}+\frac{{\beta }^2}{b^4}=\frac{1}{k^2}$
Therefore, locus is $\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{k^2}.$