MCQ
Area included between the two curves ${y^2} = 4ax$ and ${x^2} = 4ay,$ is
- A$\frac{{32}}{3}\,{a^2}\,\, sq. \,unit$
- B$\frac{{16}}{3}\,\, sq. \,unit$
- C$\frac{{32}}{3}\,\, sq. \,unit$
- ✓$\frac{{16}}{3}\,{a^2}\,\, sq. \,unit$
✓
Answer
Correct option: D.
$\frac{{16}}{3}\,{a^2}\,\, sq. \,unit$
d
(d) Solving the two equations, we have ${x^4} = 64{a^3}x$
(d) Solving the two equations, we have ${x^4} = 64{a^3}x$
$ \Rightarrow x = 0,\,\,4a$
Required area $=\int_0^{4a} {2{a^{1/2}}{x^{1/2}}dx - \int_0^{4a} {\frac{{{x^2}}}{{4a}}dx} } $
$ = \frac{{32}}{3}{a^2} - \frac{{16}}{3}{a^2} = \frac{{16}}{3}{a^2}\,\, sq. \,unit$.
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