Question

The area between the parabolas $y^2=4a(x+a)$ and $y^2=-4a(x-a)$ in sq. units is

• A. $\frac{4a^2}{3}$
• B. $\frac{8a^2}{3}$
• C. $\frac{12a^2}{3}$
• D. $\frac{16a^2}{3}$

Answer 1

Answer: (D)

$\frac{16a^2}{3}$

$y^2=4a(x+a)$ and $y^2=-4a(x-a)$
$4\stackrel{0}{\underset{-a}{\int }}\sqrt{4a(x+a)}dx=4\times 2\sqrt{a}\times \frac{2}{3}(x+a{)}^{3/2}{|}_{-a}^0$
$=\frac{16}{3}\sqrt{a}\left(a^{3/2}\right)$
$=\frac{16a^2}{3}squnits.$