The area of the trapezium whose vertices lie on the parabola $y^{2} = 4 x$ and diagonals pass through $(1, 0)$ is $k$ units. If the length of the diagonals is $\frac{25}{4}$ units each, then the value of $4 k$ is

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Solution

Since, the focal distance of any point on the parabola, $y^{2} = 4 a x$ is $a + a t^{2}$
$∴ A S = 1 + t^{2}$

$C S = A C - A S = \frac{25}{4} - (1 + t^{2})$
$A C$ is a focal chord.
$∴ \frac{1}{A S} + \frac{1}{C S} = \frac{1}{a}$
$\Rightarrow \frac{1}{1 + t^{2}} + \frac{4}{25 - 4 (1 + t^{2})} = 1$
$\Rightarrow 4 (1 + t^{2})^{2} - 25 (1 + t^{2}) + 25 = 0$
$\Rightarrow 1 + t^{2} = 5, \frac{5}{4}$
$\Rightarrow t = \pm 2, \pm \frac{1}{2}$

Hence, the coordinates of $A, B, C$ and $D$ are $(\frac{1}{4}, 1), (4, 4), (4, - 4)$ and $(\frac{1}{4}, - 1)$ respectively.

$∴ A D = 2, B C = 8$ and the distance between $A D$ and $B C$ is $\frac{15}{4}$ .

Hence, area of trapezium $A B C D$ is,
$k = \frac{1}{2} (2 + 8) \times \frac{15}{4} = \frac{75}{4} sq. units$
$\Rightarrow 4 k = 75$

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