Question

Two tangents to the ellipse $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ make angles ${\theta }_1,{\theta }_2$ with the major axis. If $\tan {\theta }_1,\tan {\theta }_2$ is a constant ${}^{\prime }{k}^{\prime }$ then the equation to the locus of their point of intersection is

• A. $y^2-b^2=kxy$
• B. $y^2+b^2=kxy$
• C. $y^2-b^2=k(x^2-a^2)$
• D. $y^2+b^2=k(x^2+a^2)$

Answer 1

The correct option is A $y^2-b^2=kxy$

The condition is not clearly specified in the question. Let us assume the condition as $cot{\theta }_1+cot{\theta }_2=k$

Given equation of the ellipse is $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Equations of the tangents to this ellipse are given by,

$y=mx\pm \sqrt{a^2{m}^2+b^2}$

Let coordinates of point of intersection of these tangents are $(h,k)$

Thus, this point must satisfy equations of both the tangents

Thus, put $x=h$ and $y=k$ in above equations, we get,

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

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

Squaring both sides, we get,

$\therefore {(k-mh)}^2=a^2{m}^2+b^2$

$\therefore k^2-2mkh+m^2{h}^2=a^2{m}^2+b^2$

$\therefore a^2{m}^2-m^2{h}^2+2mkh+b^2-k^2=0$

$\therefore (a^2-h^2)+\left(2kh\right)m+(b^2-k^2)=0$

This is a quadratic equation in m.

Now, given $cot{\theta }_1+cot{\theta }_2=k^{\prime }$

$\therefore \frac{1}{tan{\theta }_1}+\frac{1}{tan{\theta }_2}=k^{\prime }$

$\therefore \frac{1}{m_1}+\frac{1}{m_2}=k^{\prime }$

$\therefore \frac{m_1+m_2}{m_1{m}_2}=k^{\prime }$

$\therefore \frac{sumofroots}{productofroots}=k^{\prime }$

$\therefore \frac{\left(\frac{-2kh}{a^2-h^2}\right)}{\left(\frac{b^2-k^2}{a^2-h^2}\right)}=k^{\prime }$

$\therefore \frac{-2kh}{b^2-k^2}=k^{\prime }$

$\therefore -2kh=k^{\prime }(b^2-k^2)$

Replace $h$ by $x$ and $k$ by $y$, we get,

$\therefore -2yx=k^{\prime }(b^2-y^2)$

$\therefore 2yx=k^{\prime }(y^2-b^2)$