Question

If the chord of contact of the tangents drawn from the points $(\alpha, \beta)$ to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ touches the circles $x^{2}+y^{2}=c^{2}$, then the locus of the point $(\alpha, \beta)$ is

• A. $\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=\frac{1}{ c ^{2}}$
• B. $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=\frac{1}{c^{4}}$
• C. $\frac{ x ^{2}}{ a ^{4}}+\frac{ y ^{2}}{ b ^{4}}=\frac{1}{ c ^{2}}$
• D. None of these

Answer 1

Answer: (C)

Let $(h, k)$ be the point then the equation of chord of contact for the ellipse
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is
$\frac{ hx }{ a ^{2}}+\frac{ k }{ b ^{2} y }-1=0$
Since it touch the circle $x^{2}+y^{2}=c^{2}$.
$\therefore \frac{(-1)^{2}}{\frac{ h ^{2}}{ a ^{4}}+\frac{ k ^{2}}{ b ^{4}}}= c ^{2} $
$ \Rightarrow \frac{ x ^{2}}{ a ^{4}}+\frac{ y ^{2}}{ b ^{4}}=\frac{1}{ c ^{2}}$