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 in $R$. If $\theta_{1},\theta_{2}$ and $\phi$ are the angles made with the axis by the tangents at $P$ and $Q$ on the parabola and by $OR$, then the value of, $cot\theta_{1} + cot\theta_{2}$ is equal to:
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Solution
Sol. [Key Concept: On $y^{2} = 4ax$ the tangent at $t$ has $\cot\theta = t$; the circles on $OP,OQ$ as diameters meet at the foot $R$ of the perpendicular from $O$, giving $\cot\theta_{1} + cot\theta_{2} = - 2tan\varphi$.]
$\angle ORP = \frac{\pi}{2} = \angle ORQ$ slope of $PQ = \frac{2}{t_{1} + t_{2}}$
⇒ slope of $OR = - \left( \frac{t_{1} + t_{2}}{2} \right) = tan\phi$
$${\Rightarrow \frac{1}{t_{1}} = tan\theta_{1}\ \frac{1}{t_{2}} = tan\theta_{2} }{\Rightarrow tan\phi = - \left( \frac{cot\theta_{1} + cot\theta_{2}}{2} \right) }{\Rightarrow cot\theta_{1} + cot\theta_{2} = - 2tan\phi }$$
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