Question

The area included between the parabolas $y^2=4a(x+a)$ and $y^2=4b(x-a),b>a>0$, 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 $y^2=4a(x+a)...(i)$, a parabola with vertex $(-a,0)$
and $y^2=4b(x-a)...(ii)$, a parabola with vertex $(a,0)$
Solving (i) and (ii), we get $y=\pm a\sqrt{\frac{8b}{b-a}}$

$A=2\underset{0}{\overset{\frac{a\sqrt{8b}}{\sqrt{b-a}}}{\int }}\left(\left(\frac{y^2}{4b}+a\right)-\left(\frac{y^2}{4b}-a\right)\right)dy$
$=2\underset{0}{\overset{\frac{a\sqrt{8b}}{\sqrt{b-a}}}{\int }}\left(2a-\frac{\left(b-a\right)y^2}{4ab}\right)dy$
$=2{\left[2ay-\frac{b-a}{12ab}{y}^3\right]}_0^{\frac{a\sqrt{8b}}{\sqrt{b-a}}}$
$=2\left[2a\times \frac{a\sqrt{8b}}{\sqrt{b-a}}-\frac{b-a}{12ab}{\left(\frac{a\sqrt{8b}}{b-a}\right)}^3\right]$
$=4a^2\sqrt{\frac{8b}{b-a}}-\frac{4}{3}{a}^2\sqrt{\frac{8b}{b-a}}=\frac{8a^2}{3}\sqrt{\frac{8b}{b-a}}$ sq. units