Question

Through the vertex $O$ of the parabola $y^2=4ax,$ two chords $OP$ and $OQ$ are drawn and the circles on $OP$ and $OQ$ as diameters intersect at $R.$ If ${\theta }_1,{\theta }_2$ and $\varphi$ are the inclinations of the tangents at $P$ and $Q$ on the parabola and of $OR$ respectively, then the value of $\cot {\theta }_1+\cot {\theta }_2$ is

• A. $-2\tan \varphi$
• B. $-2\tan (\pi -\varphi )$
• C. $0$
• D. $2\cot \varphi$

Answer 1

The correct option is A $-2\tan \varphi$


Slope of tangent at $P$ is $\frac{1}{t_1}$
$\Rightarrow \cot {\theta }_1=t_1$
Slope of tangent at $Q$ is $\frac{1}{t_2}$
$\Rightarrow \cot {\theta }_2=t_2$
Slope of $PQ$ is $\frac{2}{t_1+t_2}$
$\Rightarrow$ Slope of $OR$ is $-\frac{t_1+t_2}{2}$
$\Rightarrow \tan \varphi =-\frac{t_1+t_2}{2}$
$\Rightarrow \tan \varphi =-\frac{1}{2}(\cot {\theta }_1+\cot {\theta }_2)$
$\therefore \cot {\theta }_1+\cot {\theta }_2=-2\tan \varphi$