Question

A circle is descirbed whose centre is the vertex and whose diameter is three-quarters of the latus rectum of the parabola y² = 4ax. If PQ is the common-chord of the circle and the parabola and L₁ L₂ is the latus rectum, then the area of the trapezium PL₁ L₂Q is

(a)   (2 + √2/2)a²

(b)    4a²

(c)    2√2a²

(d)  3√2a²

Answer 1

Correct option   (a) (2 + √2/2)a²

Explanation:

Centre of circle (0, 0)

diameter = 3/4. 4a = 3a

Eqⁿ of circle x² + y²  = 9a²/4  ...(i)

Eqⁿ of parabola y² = 4ax .....(2)

coordinates of P and Q, we get after solving (1) and (2)

x² +  4ax = 9a²/4

(x + 2a)²  = (5a/2)²  

⇒ x = -2a ± 5a/2

x = a/2, -9a/2 (not possible)

p(a/2,√2a), Q(a/2, -√2a)  y = ±√2a

PQ = √2a,  L₁L₂ = 4a

Area of trapezium = 1/2(PQ + L₁L₂)x distance between them

= 1/2 (2√2a + 4a)x(a - a/2

= (√2 + 2/2)a²