The area of the loop of the curve `a y^2=x^2(a-x)` is `4a^2s qdotu n i t s` (b) `(8a^2)/(15)s qdotu n i t s` `(16 a^2)/9s qdotu n i t s` (d) None of these

Options

A. `4a^(2)` sq. units
B. `(8a^(2))/(15)` sq. units
C. `(16a^(2))/(9)` sq. units
D. None of these

Answer

B

Explanation

`ay^(2)=x^(2)(a-x)or y=pm xsqrt((a-x)/(a))`
`"Curve tracing : "y=xsqrt((a-x)/(a))`
We must have `xlea`
`"For "0ltxlea, ygt0 and" for "xlt0, ylt0`
`"Also "y=0rArrx=0,a`
Curve is symmetrical about x-axis.
`"When "xrarr-oo,yrarr-oo`
Also, it can be verified that y has only one point of maxima for `0ltxlta.`

`"Area "=2overset(a)underset(0)intxsqrt((a-x)/(a))dx`
`sqrt((a-x)/(a))=t`
`rArr" "1-(x)/(a)=t^(2)or x=a (1-t^(2))`
`rArr" "A=2overset(0)underset(1)inta(1-t^(2))t(-2at)dt`
`4a^(2)overset(1)underset(0)int(t^(2)-t^(4))dt`
`=4a^(2)[(t^(3))/(3)-(t^(5))/(5)]_(0)^(1)`
`=4a^(2)[(1)/(3)-(1)/(5)]`
`=(8a^(2))/(15)` sq. units.