Question

If two tangents to $y^2=4ax$ make angles ${\theta }_1,{\theta }_2$ with positive $x-$ axis such that $\cos {\theta }_1\cdot \cos {\theta }_2=k,$ then the locus of their point of intersection is

• A. $x^2=k^2\left[\right(x-a{)}^2+y^2]$
• B. $x^2=k^2\left[\right(x+a{)}^2+y^2]$
• C. $x^2=k^2\left[\right(x-a{)}^2-y^2]$
• D. $x^2=\left[\right(x+a{)}^2+y^2]$

Answer 1

The correct option is A $x^2=k^2\left[\right(x-a{)}^2+y^2]$

Equation of tangent
$y=mx+\frac{a}{m}\Rightarrow m^{2}x-my+a=0$
So,
$m_1+m_2=\tan {\theta }_1+\tan {\theta }_2=\frac{y}{x}{m}_1\cdot m_2=\tan {\theta }_1\cdot \tan {\theta }_2=\frac{a}{x}$
Now,
$\cos {\theta }_1\cdot \cos {\theta }_2=k\Rightarrow \sec {\theta }_1\cdot \sec {\theta }_2=\frac{1}{k}\Rightarrow {\sec }^2{\theta }_1\cdot {\sec }^2{\theta }_2=\frac{1}{k^2}\Rightarrow (1+{\tan }^2{\theta }_1)\cdot (1+{\tan }^2{\theta }_2)=\frac{1}{k^2}\Rightarrow [1+(\tan {\theta }_1+\tan {\theta }_2{)}^2-2\tan {\theta }_1\tan {\theta }_2+{\tan }^2{\theta }_1\cdot {\tan }^2{\theta }_2]=\frac{1}{k^2}$

Let the required point be $(x_1,y_1)$, so
$[1+\frac{y_1^2}{x_1^2}-\frac{2a}{x_1}+\frac{a^2}{x_1^2}]=\frac{1}{k^2}\Rightarrow k^2\left[\right(x_1-a{)}^2+y_1^2]=x_1^2$

So, the required locus is
$x^2=k^2\left[\right(x-a{)}^2+y^2]$