Let ABCD be a trapezium whose vertices lie on | vedclass

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(A) The parabola is $y^2=4ax$ with $a=1$. Let the coordinates of $A$ be $(t_1^2, 2t_1)$ and $C$ be $(t_2^2, 2t_2)$. Since $AD$ and $BC$ are parallel t

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$\frac{75}{4}$

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(A) The parabola is $y^2=4ax$ with $a=1$. Let the coordinates of $A$ be $(t_1^2, 2t_1)$ and $C$ be $(t_2^2, 2t_2)$. Since $AD$ and $BC$ are parallel to the $y$-axis,$A$ and $D$ have the same $x$-coordinate,and $B$ and $C$ have the same $x$-coordinate. Thus,$D$ is $(t_1^2, -2t_1)$ and $B$ is $(t_2^2, -2t_2)$.Since the diagonal $AC$ passes through the focus $S(1,0)$,the points $A, S, C$ are collinear. For a chord passing through the focus with parameters $t_1$ and $t_2$,we have $t_1 t_2 = -1$,so $t_2 = -\frac{1}{t_1}$.The coordinates are $A(t_1^2, 2t_1)$ and $C(\frac{1}{t_1^2}, -\frac{2}{t_1})$.The length of $AC$ is given by $\sqrt{(t_1^2 - \frac{1}{t_1^2})^2 + (2t_1 + \frac{2}{t_1})^2} = \frac{25}{4}$.Simplifying,$\sqrt{(t_1 - \frac{1}{t_1})^2 (t_1 + \frac{1}{t_1})^2 + 4(t_1 + \frac{1}{t_1})^2} = |t_1 + \frac{1}{t_1}| \sqrt{(t_1 - \frac{1}{t_1})^2 + 4} = |t_1 + \frac{1}{t_1}| \sqrt{t_1^2 - 2 + \frac{1}{t_1^2} + 4} = (t_1 + \frac{1}{t_1})^2 = \frac{25}{4}$.Thus,$t_1 + \frac{1}{t_1} = \frac{5}{2}$,which gives $t_1 = 2$ or $t_1 = \frac{1}{2}$.Taking $t_1 = 2$,we get $A(4, 4)$ and $D(4, -4)$. Then $t_2 = -\frac{1}{2}$,so $C(\frac{1}{4}, -1)$ and $B(\frac{1}{4}, 1)$.The parallel sides are $AD = 8$ and $BC = 2$. The height of the trapezium is the difference in $x$-coordinates: $h = 4 - \frac{1}{4} = \frac{15}{4}$.Area $= \frac{1}{2} (AD + BC) 	imes h = \frac{1}{2} (8 + 2) 	imes \frac{15}{4} = \frac{1}{2} 	imes 10 	imes \frac{15}{4} = \frac{75}{4}$.

Let $ABCD$ be a trapezium whose vertices lie on the parabola $y^2=4x$. Let the sides $AD$ and $BC$ of the trapezium be parallel to the $y$-axis. If the diagonal $AC$ is of length $\frac{25}{4}$ and it passes through the point $(1,0)$,then the area of $ABCD$ is:

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