Through the vertex O of the parabola `y^(2) = 4ax`, a perpendicular is drawn to any tangent meeting it at P and the parabola at Q. Then OP, 2a and OQ are in

Options

A. (a) A.P.
B. (b) G.P.
C. (c) H.P.
D. (d) none of these

Answer

B

Explanation

Any tangent to the parabola is
`y = mx + (a)/(m)` (i)
Any line through vetex O and perpendicular to tangent is
`y =- (1)/(m)x` (ii)
OP is perpendicular distance of `O(0,0)` from (i).
`:. OP = (a)/(sqrt(1+m^(2))).(1)/(m)`
The line (ii) meets the parabola `y^(2) = 4ax` at Q.
`:.(-(1)/(m)x^(2)) =4ax`
`rArr x = 4am^(2)` and `y =- 4am`
So, Q is `(4am^(2),-4am)`.
`:. OQ = 4am sqrt((m^(2)+1))` (using distance formula)
`:. OP xx OQ = ((a)/(sqrt(1+m^(2))).(1)/(m)).4amsqrt(1+m^(2)) = 4a^(2)`
Thus, `OP, 2a, OQ` are in G.P.