MCQ
The area between the parabola ${y^2} = 4ax$ and ${x^2} = 8ay$ is
- A$\frac{8}{3}{a^2}$
- B$\frac{4}{3}{a^2}$
- ✓$\frac{{32}}{3}{a^2}$
- D$\frac{{16}}{3}{a^2}$
✓
Answer
Correct option: C.
$\frac{{32}}{3}{a^2}$
c
(c) Required area $A = \int_0^{(a{2^{8/3}})} {\sqrt {4ax} } dx - \int_0^{a{2^{8/3}}} {\frac{{{x^2}}}{{8a}}} \,dx = \frac{{32{a^2}}}{3}$
(c) Required area $A = \int_0^{(a{2^{8/3}})} {\sqrt {4ax} } dx - \int_0^{a{2^{8/3}}} {\frac{{{x^2}}}{{8a}}} \,dx = \frac{{32{a^2}}}{3}$
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