Question

Show that the area enclosed between the curve y2 = 12(x + 3) and y2 = 20(5 – x) is 64 √(5/3).

Answer 1

Equation of the curves are

y2 = 12 (x + 3) —— (1)

y2 = 20 (5 – x) —— (2)

Eliminating y

12(x + 3) = 20(5 – x)

3x + 9 = 25 – 5x

8x = 16

x = 2

y2 = 12(2 + 3) = 60

y = √60 = ± 2√15

Points of intersection are B'(2, 2√15)

B' (+2, − 2√15)

The required area is symmetrical about X – axis

Area ABCB'