Through the vertex O of the parabola $y^{2} = 4 ax$ ,two chords OP and OQ are drawn and the circles on OP and OQ as diameters intersect at R. If $θ_{1}, θ_{2}$ , and $ϕ$ ate the angles made with the axis by the tangents at P and Q on the parabola and by OR, then the value of cot $θ_{1} + \cot θ_{2}$ is

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$- 2 \tan ϕ$

$- 2 \tan (π - ϕ)$

$2 \cot ϕ$

answer is B.

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Detailed Solution

The slope of tangent at P is $1 / t_{1}$
and that at Q is $1 / t_{2}$ .Therefore,
$\cot θ_{1} = t_{1}$ and $\cot θ_{2} = t_{2}$
Slope of PQ $= \frac{2}{t_{1} + t_{2}}$
$∴$ Slope of OR $= - \frac{t_{1} + t_{2}}{2} = \tan ϕ$
(Angle in a semicircle is 90°)
or $\tan ϕ = - \frac{1}{2} (\cot θ_{1} + \cot θ_{2})$
$∴ \cot θ_{1} + \cot θ_{2} = - 2 \tan ϕ$