Question

The tangents drawn from a point $P$ to the ellipse make angles ${\theta }_1$ and ${\theta }_2$ with the major axis; find the locus of $P$ when ${\theta }_1+{\theta }_2$ is constant $(=2a)$.

Answer 1

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

Let point $P$ be $(h,k)$

Equation of tangent in slope form is $y=mx\pm \sqrt{a^2{m}^2+b^2}$

It passes through $(h,k)$

$k=mh\pm \sqrt{a^2{m}^2+b^2}k-mh=\pm \sqrt{a^2{m}^2+b^2}$

Squaring both sides

$k^2+m^2{h}^2-2hkm=a^2{m}^2+b^2(h^2-a^2)m^2-2hkm+k^2-b^2=0m_1+m_2=-\frac{b}{a}=\frac{2hk}{h^2-a^2}........\left(i\right)m_1{m}_2=\frac{c}{a}=\frac{k^2-b^2}{h^2-a^2}.........\left(ii\right){\theta }_1+{\theta }_2=2a\tan ({\theta }_1+{\theta }_2)=\tan 2a\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}=\tan 2a\frac{m_1+m_2}{1-m_1{m}_2}=\tan 2a$

SUbstituting $\left(i\right)$ and $\left(ii\right)$
$\frac{\frac{2hk}{h^2-a^2}}{1-\frac{k^2-b^2}{h^2-a^2}}=\tan 2a\frac{\frac{2hk}{h^2-a^2}}{\frac{h^2-a^2-k^2+b^2}{h^2-a^2}}=\tan 2a\frac{2hk}{h^2-a^2-k^2+b^2}=\frac{1}{\cot 2a}{h}^2-k^2+b^2-a^2=2hk\cot 2a{h}^2-k^2-2hk\cot 2a=a^2-b^2$

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

$x^2-y^2-2xy\cot 2a=a^2-b^2$

is the required locus.