Question

Find the area bounded by the parabola y2 = 4ax and its latus rectum.

Answer 1

Given parabola is $y^2 = 4ax$

 Its focus is $(a, 0)$

 Equation of latus rectum is $x = a$

The parabola is symmetrical about the x-axis  

Required area = $2[\text{ Area in the first quadrant between the limits } x = 0 \text{ and } x = a]$

$=2\int_0^aydx\\[9pt]=2\int_0^a\sqrt{4ax}dx\\[9pt]=2(2\sqrt{a})\int_0^ax^{\frac{1}{2}}dx \\[9pt]=4\sqrt{a}[\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}]|_0^a \\[9pt]=4\sqrt{a}(\dfrac{8}{3}a^2)=\dfrac{8}{3}a^2 sq. units$