Question

Find the locus of the intersection of tangents of ellipse if the lines joining the points of contact to the centre be perpendicular.

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\mathrm{1.....}\left(i\right)$

Let tangents are drawn from point $(h,k)$

Equation of chord is $T=0$

$\frac{x{x}^{\prime }}{a^2}+\frac{y{y}^{\prime }}{b^2}-1=0\frac{xh}{a^2}+\frac{yk}{b^2}=1$

Making the equation of ellipse homogenous using the equation of chord

$\frac{x^2}{a^2}+\frac{y^2}{b^2}={(\frac{xh}{a^2}+\frac{yk}{b^2})}^2\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{x^2{h}^2}{a^4}+\frac{y^2{k}^2}{b^4}+\frac{hkxy}{a^4{b}^4}{x}^2(\frac{1}{a^2}-\frac{h^2}{a^4})+y^2(\frac{1}{b^2}-\frac{k^2}{b^4})-\frac{hkxy}{a^4{b}^4}=0$

The pair of line is perpendicular

$\therefore a+b=0(\frac{1}{a^2}-\frac{h^2}{a^4})+(\frac{1}{b^2}-\frac{k^2}{b^4})=0\frac{a^2-h^2}{a^4}+\frac{b^2-k^2}{b^4}=0a^2{b}^4+b^2{a}^4=b^4{h}^2+a^4{k}^2{b}^4{h}^2+a^4{k}^2=a^2{b}^2(a^2+b^2)$

Replacing $h$ by $x$ and $y$ by $k$

$b^4{x}^2+a^4{y}^2=a^2{b}^2(a^2+b^2)$

is the required equation of locus.