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 $(=c)$.

Answer 1

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

Let the 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

$k^2+m^2{h}^2-2hkm=a^2{m}^2+b^2(h^2-a^2)m^2-2hkm+k^2-b^2=0$

Sum of roots $=m_1+m_2=-\frac{b}{a}=\frac{2hk}{h^2-a^2}........\left(i\right)$

$\tan {\theta }_1+\tan {\theta }_2=c{m}_1+m_2=c$

Substituting $\left(i\right)$

$\frac{2hk}{h^2-a^2}=cc{h}^2-c{a}^2=2hkc{h}^2-2hk=c{a}^2$

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

$c{x}^2-2xy=c{a}^2$

is the required locus.