question_answer

The area of the region bounded by the parabola\[{{y}^{2}}=4\text{ a}x\]and its latusrectum is

A)  \[\frac{4a}{3}sq\,unit\]

B)  \[\frac{4{{a}^{3}}}{3}sq\,unit\]

C)  \[\frac{8a}{3}sq\,unit\]

D)  \[\frac{8{{a}^{2}}}{3}sq\,unit\]

Correct Answer: D

Solution :

 Given, parabola is\[{{y}^{2}}=4\text{ }ax\] Required area\[=2\int_{0}^{a}{\sqrt{4ax}}\,dx\] \[=4\sqrt{a}\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{a}\] \[=4\sqrt{a}\frac{2}{3}[{{a}^{3/2}}]\] \[=\frac{8}{3}{{a}^{(3/2)+(1/2)}}=\frac{8}{3}{{a}^{2}}\]sq units