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Equation of Normal at Given Point

A chord of a ...

Question

$A$ chord of a parabola cuts the axis of the parabola at $O$ . The feet of the perpendiculars from $P$ and $P^{'}$ on the axis are $M$ and $M^{'}$ respectively. lf $V$ is the vertex then, $V M, V O, V M^{'}$ are is

A

A.P.

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B

G.P

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C

H.P.

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D

A.G..P

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Solution

The correct option is B G.P
Let parabola is $y^{2} = 4 a x$ and vertex $V = (0, 0)$
The chord $P P^{'}$ cuts axis at $O$ let it be $(k, o)$
Let $P = (a t_{1}^{2}, 2 a t_{1})$ foot of perpendicular on axis is $M = (a t_{1}^{2}, 0)$
Similarly for $P^{'} (a t_{2}^{2}, 2 a t_{2})$ foot of perpendicular is $M^{'} (a t_{2}^{2}, 0)$
Slope of $P O = \frac{2 a t_{1}}{a t_{1}^{2} - k}$ must be equal to $O P^{'} = \frac{- 2 a t_{2}}{k - a t_{2}^{2}}$
Therefore $\frac{2 a t_{1}}{a t_{1}^{2} - k} = \frac{- 2 a t_{2}}{k - a t_{2}^{2}}$
Solving this we get $k = - a t_{1} t_{2}$
$V O = k, V M = a t_{1}^{2}, V M^{'} = a t_{2}^{2}$
$V O^{2} = k^{2} = (- a t_{1} t_{2})^{2} = a t_{1}^{2} a t_{2}^{2} = V M * V M^{'}$
Which gives $V M, V O, V M^{'}$ in G.P

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