Q. A trapesium is inscribed in the parabola $y^{2}-4 x$ such that its diagonals passes through $M(1,0)$ and each has length $\frac{25}{4}$, The area of trapezium is equal to
Conic Sections
Solution:
The diagonals are focal choed.
$As , 1+ t ^{2}- c (say)$
Now, $\frac{1}{c}+\frac{1}{\left(\frac{25}{4}-c\right)}=1 \left(\Theta \frac{1}{AS}+\frac{1}{CS}=\frac{1}{a}\right)$
$\Rightarrow c -\frac{5}{4}, 5$
$\therefore A \left(\frac{1}{4}, 1\right), B (4,4), C (4,-4)$ and $D \left(\frac{1}{4},-1\right)$
So, area of trapezium $-\frac{1}{2}(2+8) \times \frac{15}{4}-\frac{75}{4}$.
