question_answer

The area cut off by the latus rectum from the parabola \[{{y}^{2}}=4ax\] is:

A) \[(8/3)a\,\,sq\,\,unit\] 

B) \[(8/3)\sqrt{a}\,\,sq\,\,unit\]

C) \[(3/8)\sqrt{a}\,\,sq\,\,unit\]

D) \[(8/3){{a}^{2}}\,\,sq\,\,unit\]

Correct Answer: D

Solution :

\[\therefore \]Required area\[=2\int_{0}^{a}{y}\,dx\]                                 \[=2\int_{0}^{a}{2\sqrt{a}\sqrt{x}dx}\]                 \[=4\sqrt{a}\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{a}\]                 \[=\frac{8}{3}{{a}^{2}}\,\,ssq\,\,unit\]