Let ABCD be a trapezium whose vertices lie on the parabola y^2=4x.

Question

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 $\frac{25}{4}$ and it passes through the point $(1,0)$, then the area of $ABCD$ is

Accepted Answer

Focus $(1,0)$; $AC$ is a focal chord, $A=(t_1^2,2t_1),\ C=\left(\dfrac{1}{t_1^2},-\dfrac{2}{t_1}\right)$.

$$AC=\left(t_1+\frac1{t_1}\right)^2=\frac{25}{4} \;\Rightarrow\; t_1+\frac1{t_1}=\pm\frac52 \;\Rightarrow\; t_1=2\text{ or }\frac12.$$

Taking $t_1=2$: $A=(4,4)$, $C=\left(\dfrac14,-1\right)$. With $AD,BC\parallel y$-axis, $D=(4,-4)$ and $B=\left(\dfrac14,1\right)$.

Parallel sides $AD=8$ (at $x=4$) and $BC=2$ (at $x=1/4$); height $=4-\dfrac14=\dfrac{15}{4}$.

$$\text{Area}=\frac12(8+2)\left(\frac{15}{4}\right)=\frac{75}{4}.$$

Solutions · 1