Question

Find the locus of the middle points of chords of an ellipse the tangents at the ends of which intersect at right angles.

Answer 1

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

Let $(h,k)$ be the mid point of chord

Equation of chord with given mid point is $T=S^{\prime }$

$\frac{\frac{xh}{a^2}}{(\frac{h^2}{a^2}+\frac{k^2}{b^2})}+\frac{\frac{yk}{b^2}}{(\frac{h^2}{a^2}+\frac{k^2}{b^2})}=1\frac{\frac{xh}{a^2}}{\frac{b^2{h}^2+a^2{k}^2}{a^2{b}^2}}+\frac{\frac{yk}{b^2}}{\frac{b^2{h}^2+a^2{k}^2}{a^2{b}^2}}=1\frac{b^{2}hx}{b^2{h}^2+a^2{k}^2}+\frac{a^{2}ky}{b^2{h}^2+a^2{k}^2}=\mathrm{1.........}\left(i\right)$

Let the point of intersection of the tangents at the end points of the chord be $P(p,q)$

Equation of chord of contact w.r.t to a given point is $T=0$

$\frac{x{x}^{\prime }}{a^2}+\frac{y{y}^{\prime }}{b^2}-1=0\frac{xp}{a^2}+\frac{yq}{b^2}=\mathrm{1......}\left(ii\right)$

$\left(i\right)$ and $\left(ii\right)$ represents the same line, so by comparing both the lines

$\frac{p}{a^2}=\frac{b^{2}h}{b^2{h}^2+a^2{k}^2}\Rightarrow p=\frac{{a^{2}b}^{2}h}{b^2{h}^2+a^2{k}^2}\frac{q}{b^2}=\frac{a^{2}k}{b^2{h}^2+a^2{k}^2}\Rightarrow q=\frac{{a^{2}b}^{2}k}{b^2{h}^2+a^2{k}^2}$

So point $P$ is $(\frac{{a^{2}b}^{2}h}{b^2{h}^2+a^2{k}^2},\frac{{a^{2}b}^{2}k}{b^2{h}^2+a^2{k}^2})$

Both the tangents are perpendicular their point of intersection i.e. $P$ will lie on the director circle of the ellipse.

$x^2+y^2=a^2+b^2{\left(\frac{{a^{2}b}^{2}h}{b^2{h}^2+a^2{k}^2}\right)}^2+{\left(\frac{{a^{2}b}^{2}k}{b^2{h}^2+a^2{k}^2}\right)}^2=a^2+b^2{a^{4}b}^4(h^2+k^2)=(a^2+b^2){(b^2{h}^2+a^2{k}^2)}^2$

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

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

is the required equation of locus.