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Q.
The area included between the parabolas $y^2 = 4a (x + a)$ and $y^2 = 4b(x - a), b > a > 0$, is
Application of Integrals
Solution:
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 $y = \pm a\sqrt{\frac{8b}{b-a}}$
$A = 2 \int\limits^{\frac{a\sqrt{8b}}{\sqrt{b-a}}}_{0} \left(\left(\frac{y^{2}}{4b}+a\right)-\left(\frac{y^{2}}{4b}-a\right)\right)dy$
$= 2 \int\limits^{\frac{a\sqrt{8b}}{\sqrt{b-a}}}_{0} \left(2a-\frac{\left(b-a\right)y^{2}}{4ab}\right)dy$
$= 2\left[2ay-\frac{b-a}{12ab}y^{3}\right]^{\frac{a\sqrt{8b}}{\sqrt{b-a}}}_{0}$
$= 2\left[2a\times \frac{a\sqrt{8b}}{\sqrt{b-a}}-\frac{b-a}{12ab}\left(\frac{a\sqrt{8b}}{b-a}\right)^{3}\right]$
$=4a^{2}\sqrt{\frac{8b}{b-a}} -\frac{4}{3}a^{2}\sqrt{\frac{8b}{b-a}} = \frac{8a^{2}}{3}\sqrt{\frac{8b}{b-a}}$ sq. units
