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 $y$-axis. If the diagonal $AC$ is of length $\tfrac{25}{4}$ and it passes through the point $(1, 0)$, then the area of $ABCD$ is
Step-by-Step Solution
Key Concept: Use the focal chord property: $AC$ passes through the focus $(1,0)$ of $y^2=4x$, so $AC$ is a focal chord; its length $a(t_1+1/t_1)^2=25/4$ gives $t_1=2$ or $1/2$; then identify all four vertices and apply the trapezium area formula.
The point $(1,0)$ is the focus of $y^2=4x$ ($a=1$). So $AC$ is a focal chord with endpoints $A=(t_1^2,2t_1)$ and $C=(1/t_1^2,-2/t_1)$. Focal chord length $=(t_1+1/t_1)^2=25/4$, so $t_1+1/t_1=\pm\tfrac{5}{2}$, giving $t_1=2$ or $\tfrac{1}{2}$. Taking $t_1=\tfrac{1}{2}$: $A=(\tfrac{1}{4},1)$, $C=(4,-1)$. Then $B=(4,4)$, $D=(\tfrac{1}{4},-1)$... wait: $AD\parallel y$-axis so $A,D$ share $x=\tfrac{1}{4}$, i.e., $D=(\tfrac{1}{4},-1)$; $BC\parallel y$-axis so $B,C$ share $x=4$, giving $B=(4,4)$. Lengths: $|AD|=1+1=2$ ... actually $A=(\tfrac{1}{4},1)$, $D=(\tfrac{1}{4},-1)$: $|AD|=2$. $B=(4,4)$, $C=(4,-4)$: $|BC|=8$. Height (distance between parallel sides) $=4-\tfrac{1}{4}=\tfrac{15}{4}$. Area $=\tfrac{1}{2}(2+8)\cdot\tfrac{15}{4}=\tfrac{75}{4}$.
Correct Answer: 1