If $P, Q, R$ are three points on a parabola $y^{2} = 4 a x$ whose ordinates are in geometrical progression, then the tangents at $P$ and $R$ meet on

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the line through Q parallel to y-axis

the line through Q parallel to x-axis

the line joining Q to the focus.

the line joining Q to the vertex

answer is B.

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Detailed Solution

Let the coordinates of $P, Q, R$ be $(a t^{2}, 2 a t) i = 1, 2, 3$ having ordinates in G.P. So that $t_{1}, t_{2}, t_{3}$ are also in

G.P. i.e. $t_{1} t_{3} = {t_{2}}^{2}$ . Equations of the tangents at P and R are $t_{1} y = x + a t_{1}^{2}$ and $t_{3} y = x + a t_{3}^{2},$ which intersect at the point $\frac{x + a t_{1}^{2}}{t_{1}} = \frac{x + a t_{3}^{2}}{t_{3}} \Rightarrow x = a t_{1} t_{3} = a t_{2}^{2}$ which is a line through $Q$ parallel to y-axis.