Question

The area cut off by the latus rectum from the parabola $y^2=4ax$

• A. $\frac{8}{3}a$
• B. $\frac{8}{3}\sqrt{a}$
• C. $\frac{8}{3}{a}^2$
• D. $\frac{8}{3}{a}^3$

Answer 1

The correct option is C $\frac{8}{3}{a}^2$

For parabola,

$y^2=4ax$

Latus Rectum is the line $x=a$

Area required$=$Area OLSL'

$=2\times AreaOSL$

$=2\times {\int }_0^{a}ydx$

$=2\times {\int }_0^a\sqrt{4ax}dx$ $(\because y^2=4ax)\Rightarrow y=\sqrt{4ax})$

$=2\times {\int }_0^a\sqrt{4a}\sqrt{x}dx$

$=4\sqrt{a}{\int }_0^a\sqrt{x}dx$

$=4\sqrt{a}{\int }_0^a\sqrt{x}dx$

$=4\sqrt{a}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}$

$=\frac{8}{3}\sqrt{a}{a}^{\frac{3}{2}}$

$=\frac{8}{3}{a}^{\frac{1}{2}}{a}^{\frac{3}{2}}$

$=\frac{8}{3}{a}^{\frac{4}{2}}$

$=\frac{8}{3}{a}^2$