Question

Find the area bounded by $y^2=4ax$ and the tangents at the ends of its latus rectum.?

Answer 1

Given equation of the curve,

$y^2=4ax\Rightarrow y=2a^{\frac{1}{2}}\cdot x^{\frac{1}{2}}$

and the latus rectum is a line with equation $x=a$.

$\therefore$ Area bounded by the curve and the latus rectum

=$2\times$ Area of the first quadrant

$=2{\int }_0^{a}ydx=2{\int }_0^{a}2a^{\frac{1}{2}}\cdot x^{\frac{1}{2}}dx=4a^{\frac{1}{2}}{[\frac{x}{3}\cdot x^{\frac{3}{2}}]}_0^a=\frac{8}{3}{a}^{\frac{1}{2}}\cdot a^{\frac{3}{2}}=\frac{8}{3}{a}^2$

Hence, The area bounded by the curve $y^2=4ax$ and the tangents at the ends of its latus rectum is $\frac{8}{3}{a}^2.$