Question

Two tangents to $y^2=4ax$ make angles ${\theta }_1,{\theta }_2$ with $x$ axis. lf $\cos {\theta }_1\cos {\theta }_2=k$ then the locus of their 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. $x^2=k^2\left[\right(x-a{)}^2+4y^2]$
• D. $4x^2=k^2\left[\right(x+a{)}^2+y^2]$

Answer 1

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

equation of tangent in slope form is $m^{2}x-ym+a=0$
Therefore, $m_1+m_2=\tan {\theta }_1+\tan {\theta }_2=\frac{y}{x}$ ....(1)
and $m_1{m}_2=\tan {\theta }_1\tan {\theta }_2=\frac{a}{x}$ ....(2)
now, $\cos {\theta }_1\cos {\theta }_2=k$
$\Rightarrow {\sec }^2{\theta }_1{\sec }^2{\theta }_2=\frac{1}{k^2}$
$\Rightarrow (1+{\tan }^2{\theta }_1)(1+{\tan }^2{\theta }_2)=\frac{1}{k^2}$

$1+{(\tan {\theta }_1+\tan {\theta }_2)}^2-2\tan {\theta }_1\tan {\theta }_2+{\tan }^2{\theta }_1{\tan }^2{\theta }_2=\frac{1}{k^2}$
from (1) and (2): $1+\frac{y^2}{x^2}-\frac{2a}{x}+\frac{4a^2}{x^2}=\frac{1}{k^2}$
$x^2=k^2\left[\right(x-a{)}^2+y^2]$