A trapesium is inscribed in the parabola y2-4 x such that its diagonals passes through M(1,0) and each has length (25/4), The area of trapezium is equal to

Question

A trapesium is inscribed in the parabola y2-4 x such that its diagonals passes through M(1,0) and each has length (25/4), The area of trapezium is equal to (A) (70/4) (B) (71/4) (C) (80/4) (D) (75/4)

Tardigrade Admin · Accepted Answer

The diagonals are focal choed. As , 1+ t 2- c (say) Now, (1/c)+(1/((25)4-c))=1 (Θ (1/AS)+(1/CS)=(1/a)) ⇒ c -(5/4), 5 ∴ A ((1/4), 1), B (4,4), C (4,-4) and D ((1/4),-1) So, area of trapezium -(1/2)(2+8) × (15/4)-(75/4).

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

$\frac{70}{4}$

$\frac{71}{4}$

$\frac{80}{4}$

$\frac{75}{4}$

Q. A trapesium is inscribed in the parabola y2−4x such that its diagonals passes through M(1,0) and each has length 425​, The area of trapezium is equal to

470​

471​

480​

475​

The diagonals are focal choed. As,1+t2−c(say) Now, c1​+(425​−c)1​=1(ΘAS1​+CS1​=a1​) ⇒c−45​,5 ∴A(41​,1),B(4,4),C(4,−4) and D(41​,−1) So, area of trapezium −21​(2+8)×415​−475​.

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

$\frac{70}{4}$

$\frac{71}{4}$

$\frac{80}{4}$

$\frac{75}{4}$