Question

lf the tangents from $P$ to the circle $x^2+y^2=a^2$ make angles $\alpha$ and $\beta$ with the positive direction of the $x$-axis such that $\cot \alpha +\cot \beta =k$ then the locus of $P$ is

• A. $k{y}^2-2xy=a^2$
• B. $k(y^2-a^2)=2xy$
• C. $k(y^2-a^2)=x^2-a^2$
• D. $k(x^2-a^2)=2xy$

Answer 1

The correct option is B $k(y^2-a^2)=2xy$

$\Rightarrow$ Let the equation of tangent is
$y-k=m(x-h)$
$\Rightarrow y=mx+k-mh---\left(1\right)$
Then, $a=\left|\frac{mh-k}{\sqrt{1+m^2}}\right|$
$a^2+a^2{m}^2=m^2{h}^2-2mhk+k^2$
$(h^2-a^2)m^2-2mhk+k^2-a^2=0$
$m_1+m_2=\frac{2hk}{h^2-a^2}$
$m_1{m}_2=\frac{k^2-a^2}{h^2-a^2}$
So, $cot\alpha +cot\beta =R$
$\frac{2hk}{(k^2-a2)}=R$
$\Rightarrow$ So, $R(k^2-a^2)=2hk$
Locus of $P(h,k)$ is $R(y^2-a^2)=2xy$