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 ${\tan }^2{\theta }_1+{\tan }^2{\theta }_2$ is constant $(=\lambda )$.

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

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

${\left(\tan {\theta }_1+\tan {\theta }_2\right)}^2={\tan }^2{\theta }_1+{\tan }^2{\theta }_2+2\tan {\theta }_1\tan {\theta }_2{\left(\tan {\theta }_1+\tan {\theta }_2\right)}^2=\lambda +2\tan {\theta }_1\tan {\theta }_2{(m_1+m_2)}^2=\lambda +2m_1{m}_2$

Substituting $\left(i\right)$ and $\left(ii\right)$

${\left(\frac{2hk}{h^2-a^2}\right)}^2=\lambda +2\frac{k^2-b^2}{h^2-a^2}{\left(\frac{2hk}{h^2-a^2}\right)}^2=\frac{\lambda (h^2-a^2)+2(k^2-b^2)}{h^2-a^2}\frac{4h^2{k}^2}{h^2-a^2}=\lambda (h^2-a^2)+2(k^2-b^2)4h^2{k}^2=\lambda {(h^2-a^2)}^2+2(k^2-b^2)(h^2-a^2)4h^2{k}^2=\lambda {(h^2-a^2)}^2+2h^2{k}^2-2k^2{a}^2-2b^2{h}^2+2a^2{b}^2\lambda {(h^2-a^2)}^2=2(h^2{k}^2+k^2{a}^2+b^2{h}^2-a^2{b}^2)$

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

$\lambda {(x^2-a^2)}^2=2(x^2{y}^2+y^2{a}^2+b^2{x}^2-a^2{b}^2)$

is the required equation of locus.