Question

Tangents are drawn from a point P to the circle $x^2+y^2=r^2$ so that the chords of contact are tangents to the ellipse $a^2{x}^2+b^2+y^2=r^4$. Find the locus of P.

Answer 1

Let $P$ have coordinates $(h,k)$. The chord of contact from $P$ to the circle is $hx+ky=r^2⟹y=-\frac{h}{k}+\frac{r^2}{k}$ ... (1)

Now, $a^2{x}^2+b^2{y}^2=r^4⟹\frac{a^2{x}^2}{r^4}+\frac{b^2{y}^2}{r^4}=1⟹\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$

where $A=\frac{r^2}{a}$ and $A=\frac{r^2}{a}$ and $B=\frac{r^2}{b}$

Hence, equation of tangent to ellipse in slope form is $y=mx+\sqrt{A^2{m}^2+B^2}$ ... (2)

On comparing (1) and (2),

$m=-\frac{h}{k}$ and $\sqrt{A^2{m}^2+B^2}=\frac{r^2}{k}$

$\therefore \sqrt{\frac{r^4{h}^2}{a^2{k}^2}+\frac{r^4}{b^2}}=\frac{r^2}{k}$

Squaring both sides,

$\frac{r^4{h}^2}{a^2{k}^2}+\frac{r^4}{b^2}=\frac{r^4}{k^2}$

$\therefore \frac{h^2}{a^2{k}^2}+\frac{1}{b^2}=\frac{1}{k^2}$

$\therefore \frac{h^2}{a^2}+\frac{k^2}{b^2}=1$

Hence the required locus for point $P$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$