Question

The area included between the parabolas $y^{2}=4a \left(x+a\right)$ and $y^{2}=4b\left(x-a\right)$, $b\,>\,a>\,0$, is

• A. $\frac{4\sqrt{2}}{3}b^{2} \sqrt{\frac{a}{b-a}}$ sq. units
• B. $\frac{8\sqrt{8}}{3}a^{2} \sqrt{\frac{b}{b-a}}$ sq. units
• C. $\frac{4\sqrt{2}}{3}a^{2} \sqrt{\frac{b}{b-a}}$ sq. units
• D. $\frac{8\sqrt{8}}{3}b^{2} \sqrt{\frac{a}{b-a}}$ sq. units

Answer 1

Answer: (B)

We have, $y^{2} = 4a\left(x + a\right)$ $\,...\left(i\right)$, a parabola with vertex $\left(-a, 0\right)$
and $y^{2} = 4b\left(x - a\right)$ $\,...\left(ii\right)$, a parabola with vertex $\left(a, 0\right)$
Solving $\left(i\right)$ and $\left(ii\right)$, we get $y =\pm \,a \sqrt{\frac{8b}{b-a}}$

$A=2\int_{0}^{\frac{a\sqrt{8b}}{\sqrt{b-a}}} \left(\left(\frac{y^{2}}{4b}+a\right)-\left(\frac{y^{2}}{4a}-a\right)\right)dy$

$=2\int_{0}^{\frac{a\sqrt{8b}}{b-a}} \left(2a-\frac{\left(b-a\right)y^{2}}{4ab}\right)dy$

$=2\left[2ay-\frac{b-a}{12ab}y^{3}\right]_{0}^{\frac{a\sqrt{8b}}{\sqrt{b-a}}}$

$=2\left[2a\times\frac{a\sqrt{8b}}{\sqrt{b-a}}-\frac{b-a}{12 ab}\left(\frac{a\sqrt{8b}}{\sqrt{b-a}}\right)^{3}\right]$

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

$=\frac{8a^{2}}{3}\sqrt{\frac{8b}{b-a}}$ sq. units