Question

Find the locus of the middle points of chords of an ellipse whose poles are on the auxiliary circle.

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 when mid point is given is $T=S^{\prime }$

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

Let $(p,q)$ be pole w.r.t to this line

Equation of polar when pole is given 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 $\left(ii\right)$ are both polar w.r.t to $P$ , so by comparing both the equations

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

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

It lies on the auxiliary circle

$x^2+y^2=a^2{\left(\frac{{a^{2}b}^{2}h}{b^2{h}^2+a^2{k}^2}\right)}^2+{\left(\frac{{a^{2}b}^{2}k}{b^2{h}^2+a^2{k}^2}\right)}^2=a^2\frac{{a^{2}b}^4{h}^2}{{{(b}^2{h}^2+a^2{k}^2)}^2}+\frac{{a^{2}b}^4{k}^2}{{{(b}^2{h}^2+a^2{k}^2)}^2}=1{a^{2}b}^4(h^2+k^2)={{(b}^2{h}^2+a^2{k}^2)}^2$

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

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

is the required equation of locus.