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 $\tan {\theta }_1-\tan {\theta }_2$ is constant $(=d)$.

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

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

${\left(\tan {\theta }_1-\tan {\theta }_2\right)}^2={\left(\tan {\theta }_1+\tan {\theta }_2\right)}^2-4\tan {\theta }_1\tan {\theta }_2{d}^2={(m_1+m_2)}^2-4m_1{m}_2$

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

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

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

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

is the required equation of locus.