Question

The locus of the point of intersection of two tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which are inclined at angles ${\theta }_1$ and ${\theta }_2$ with the major axis such that ${\tan }^2{\theta }_1+{\tan }^2{\theta }_2$ is constant, is

• A. $4x^2{y}^2+2(x^2-a^2)(y^2-b^2)=k{(x^2-a^2)}^2$
• B. $4x^2{y}^2-2(x^2-a^2)(y^2-b^2)=k{(x^2-a^2)}^2$
• C. $4x^2{y}^2+2(x^2-a^2)(y^2-b^2)=k{(x^2+a^2)}^2$
• D. None of these

Answer 1

The correct option is B $4x^2{y}^2-2(x^2-a^2)(y^2-b^2)=k{(x^2-a^2)}^2$

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

$m_1+m_2=\frac{2x_1{x}_2}{{x_1}^2-a^2}$ and $m_1{m}_2=\frac{{y_1}^2-b^2}{{x_1}^2-a^2}$

Given: ${\tan }^2{\theta }_1+{\tan }^2{\theta }_2=$ constant $=k$ (say)

$\Rightarrow {m_1}^2+{m_2}^2=k\Rightarrow {(m_1+m_2)}^2-2m_1{m}_2=k$

$\Rightarrow \frac{4x^2{y}^2}{({x_1}^2-a^2)}-\frac{2({y_1}^2-b^2)}{({x_1}^2-a^2)}=k\Rightarrow 4{x_1}^2{y_1}^2-2({x_1}^2-a^2)({y_1}^2-b^2)=k{({x_1}^2-a^2)}^2$

Hence locus of $P$ is

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