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 diameter intersect in $R$. If ${\theta }_1,{\theta }_2$ and $\varphi$ are the angles made with the axis by the tangent at $P$ and $Q$ on the parabola and by $OR$ then the value of $\cot {\theta }_1+\cot {\theta }_2=$

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

Answer 1

Answer: (A)

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