Question

Find the locus of the intersection of tangents of ellipse if the sum of the ordinates of the points of contact be equal to $b$.

Answer 1

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

Let the point of intersection of tangents be P$(h,k)$

Equation of chord of contact

$\frac{hx}{a^2}+\frac{ky}{b^2}-1=\mathrm{0.....}\left(i\right)$

Let points of contact of tangent be $(a\cos \theta ,b\sin \theta )$ and $(a\cos \varphi ,b\sin \varphi )$

Equation of chord of contact

$\frac{x}{a}\cos \left(\frac{\theta +\varphi }{2}\right)+\frac{y}{b}\sin \left(\frac{\theta +\varphi }{2}\right)=\cos \left(\frac{\theta -\varphi }{2}\right)$

$\frac{x}{a}\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}+\frac{y}{b}\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}=1$ ....... $\left(ii\right)$

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

$\frac{h}{a^2}=\frac{1}{a}\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}\Rightarrow h=a\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}......\left(iii\right)\frac{k}{b^2}=\frac{1}{b}\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}\Rightarrow k=b\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\cos \left(\frac{\theta -\varphi }{2}\right)}........\left(iv\right)$

Given $b\sin \theta +b\sin \varphi =b$

$2\sin \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)=1\sin \left(\frac{\theta +\varphi }{2}\right)\cos \left(\frac{\theta -\varphi }{2}\right)=\frac{1}{2}\cos \left(\frac{\theta -\varphi }{2}\right)=\frac{1}{2}\frac{1}{\sin \left(\frac{\theta +\varphi }{2}\right)}$

Substituting in $\left(iii\right)$ and $\left(iv\right)$
$h=a\frac{\cos \left(\frac{\theta +\varphi }{2}\right)}{\frac{1}{2}\frac{1}{\sin \left(\frac{\theta +\varphi }{2}\right)}}=2a\cos \left(\frac{\theta +\varphi }{2}\right)\sin \left(\frac{\theta +\varphi }{2}\right).......\left(v\right)$

$k=b\frac{\sin \left(\frac{\theta +\varphi }{2}\right)}{\frac{1}{2}\frac{1}{\sin \left(\frac{\theta +\varphi }{2}\right)}}=2b{\sin }^2\left(\frac{\theta +\varphi }{2}\right)$

${\sin }^2\left(\frac{\theta +\varphi }{2}\right)=\frac{k}{2b}\Rightarrow \sin \left(\frac{\theta +\varphi }{2}\right)=\sqrt{\frac{k}{2b}}$

$1-{\cos }^2\left(\frac{\theta +\varphi }{2}\right)=\frac{k}{2b}{\cos }^2\left(\frac{\theta +\varphi }{2}\right)=1-\frac{k}{2b}\Rightarrow \cos \left(\frac{\theta +\varphi }{2}\right)=\sqrt{\frac{2b-k}{2b}}$

Substituting in $\left(v\right)$
$h=2a\sqrt{\frac{2b-k}{2b}}\sqrt{\frac{k}{2b}}hb=a\sqrt{2b-k}\sqrt{k}$

Squaring both sides

$h^2{b}^2=a^2(2bk-k^2)h^2{b}^2+a^2{k}^2=2a^{2}bk$

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

$x^2{b}^2+a^2{y}^2=2a^{2}by$

is the required equation of locus.