Question

If chord contact of the tangents drawn from the point $(\alpha ,\beta )$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ touches the circle $x^2+y^2=c^2,$ THEN THE LOCUS OF THE POINT $(\alpha ,\beta )$

• 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

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

Equation of chord to the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ from point $(\alpha ,\beta )$ is $\frac{x\alpha }{a^2}+\frac{y\beta }{b^2}=1-\left(1\right)$

As per question line(1) touches the circle $x^2+y^2=c^2$

Perpendicular that is distance from(0,0) from line 1 is

$\frac{0+0-1}{\sqrt{(\frac{\alpha }{a^2}{)}^2+(\frac{\beta }{b^2}{)}^2}}=c$

$⟹\frac{-1}{\sqrt{\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}}}=c$

$⟹\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}=\frac{1}{c^2}$

Locus of point ( $\alpha ,\beta )$

$⟹\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{c^2}$