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 $(\alpha ,\beta )$

Answer 1

Equation of chord to the ellipse

$\begin{array}{c}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1frompoint\left(a,b\right)\\ =\frac{x\alpha }{a^2}+\frac{y\beta }{b^2}=1\to \left(i\right)\end{array}$

line (1) touches the circle $x^2+y^2=k^2$ perpendicular that means distance (0,0) from the line (1) is

$\begin{array}{c}\frac{0\left(\frac{\alpha }{a^2}\right)+0\left(\frac{\beta }{b^2}\right)-1}{\sqrt{{\left(\frac{\alpha }{a}\right)}^2+{\left(\frac{\beta }{b}\right)}^2}}=k\\ \Rightarrow \frac{-1}{\sqrt{\frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}}}=k\\ \Rightarrow \frac{{\alpha }^2}{a^2}+\frac{{\beta }^2}{b^2}=\frac{1}{k^2}\end{array}$

locus of point $\left(\alpha \beta \right)$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{k^2}$