Question

If ${\theta }_1,{\theta }_2$ be the inclinations of tangents drawn from the point $P$ to the circle $x^2+y^2=a^2$ and $\cot {\theta }_1+\cot {\theta }_2=k$, then the locus of $P$ is

• A. $k(y^2+a^2)=2xy$
• B. $k(y^2-a^2)=2xy$
• C. $k(y^2+a^2)=4xy$
• D. none of these

Answer 1

Answer: (B)

$k(y^2-a^2)=2xy$

Equation of the circle is $x^2+y^2=a^2$ ...(1)

Let $P$ be the point $(x_1,y_1)$.

Equation of any tangent to (1) is $y=mx+a\sqrt{1+m^2}$

It is passes through $P(x_1,y_1)$, then

$y_1=m{x}_1+a\sqrt{1+m^2}\Rightarrow y_1-m{x}_1=a\sqrt{1+m^2}$

Squaring ${y_1}^2+2m{x}_1{y}_1+m^2{x_1}^2=a^2(1+m^2)$

$\Rightarrow ({x_1}^2-a^2)m^2-2x_1{y}_{1}m+({y_1}^2-a^2)=0$ ...(2)

This is a quadratic in $m$. If $m_1$ and $m_2$ are its roots, then these are the slopes of the tangents from $P$.

Since inclination of tangents are given to be ${\theta }_1$ and ${\theta }_2$

$\therefore$ Let $m_1=\tan {\theta }_1$ and $m_2=\tan {\theta }_2$

$\Rightarrow \frac{1}{m_1}+\frac{1}{m_2}=k\Rightarrow m_1+m_2=k{m}_1{m}_2$

$\therefore \frac{2x_1{y}_1}{{x_1}^2-a^2}=k.\frac{{y_1}^2-a^2}{{x_1}^2-a^2}\Rightarrow 2x_1{y}_1=k({y_1}^2-a^2)$

$\therefore$ Locus of $P$ is $k\left(y^2{-a}^2\right)=2xy$