A square has one vertex on the parabola $y^{2} = 4 a x$ and the diagonal through this vertex lies along the axis of the parabola. If the ends of the other diagonal lie on the parabola, the coordinate of one of the vertex of the square is

(0, 2 a)

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(4 a, 4 a)

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(a, - 2 a)

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none of these

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Solution

The correct option is B $(4 a, 4 a)$
Let us suppose square will be $A B C D$ ,
vertex A will be lie on origin,
coordinates of $A = (0, 0)$
coordinates of $B = (a t^{2}, 2 a t)$
coordinates of $C = (h, 0)$ . :- ( C will lie on $x - a x i s$ . )
coordinates of $D = (a t^{2}, - 2 a t)$

slope of AB $= t a n 45^{\circ} = 1$ . --(1)
we know that slope of $A B = \frac{y_{2} - y 1}{x_{2} - x 1}$
put the value of $x 1, x 2, y 1, y 2$ from coordinates A & B,

so, slope of $A B = \frac{2 a t - 0}{a t^{2} - 0}$
from equation 1, we know that,

$\frac{2 a t - 0}{a t^{2} - 0} = 1$

$2 a t = a t^{2}$
and $t = 2$

if we consider the coordinated of B,
than $B = (a t^{2}, 2 a t)$
put the value of $^{'} t^{'}$
$B = (4 a, 4 a)$

so, correct option will be 'B'

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