Question

Through the vertex $O$ of the parabola, $y^2=4 a x$ two chords $O P$ and $O Q$ are drawn and the circles on $OP$ and $OQ$ as diameter intersect in $R$. If $\theta_1, \theta_2$ and $\phi$ are the angles made with the axis by the tangent at $P$ and $Q$ on the parabola and by $O R$ then the value of $\cot \theta_1+\cot \theta_2=$

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

Answer 1

Answer: (A)

Let $P \left( at _1^2, 2 at _1\right) \& Q \left( at _2^2, 2 at _2\right)$
so $\tan \theta_1=\frac{1}{t_1} \& \tan \theta_2=\frac{1}{t_2}$
$\cot \theta_1+\cot \theta_2= t _1+ t _2$... (i)
equation of circle with $(0,0) \&\left( at _1^2, 2 at _1\right)$
as end points of diameter is
$x\left(x-a t_1^2\right)+y\left(y-2 a t_1\right) \text { so } $
$S_1: x^2+y^2-a t_1^2 x-2 a t_1 y=0$...(ii)
similarly other circle is
$S_2: x^2+y^2-a t_2^2 x-2 a t_2 y=0$ ...(iii)
equation of $OR$ will be $S _1- S _2=0$
$a\left(t_2^2-t_1^2\right) x+2 a\left(t_2-t_1\right) y=0 $
$y=-\left(\frac{t_1+t_2}{2}\right) x $
$\tan \phi=-\frac{t_1+t_2}{2}$
from (i) $\cot \theta_1+\cot \theta_2=-2 \tan \phi$