question_answer

Area bounded by curve \[{{y}^{2}}=x\] and \[x=\frac{3}{4}{{y}^{2}}+1\]is

A) \[4\]

B)        \[8/3\]

C)  \[6\]

D)         \[16/3\]

Correct Answer: B

Solution :

Given curves are, \[{{y}^{2}}=x\]                     ?..(i) \[x=\frac{3}{4}{{y}^{2}}+1\]                         ??(ii) On solving (i) and (ii), \[{{y}^{2}}=\frac{3}{4}{{y}^{2}}+1\Rightarrow \left( 1-\frac{3}{4} \right){{y}^{2}}=1\Rightarrow \frac{1}{4}{{y}^{2}}=1\] \[\Rightarrow \,\,y=\pm 2\] and \[x=4\] \[(4,2)\,(4,-2)\] are point of intersection of(i) and (ii). \[\therefore \] required area \[=2\left[ \int\limits_{0}^{2}{\left( \frac{3}{4}{{y}^{2}}+1 \right)dy-\int\limits_{0}^{2}{{{y}^{2}}dy}} \right]\] \[=2\left[ \left( \frac{3}{4}\times \frac{{{y}^{3}}}{3}+y \right)_{0}^{2}-\left( \frac{{{y}^{3}}}{3} \right)_{0}^{2} \right]\] \[=2\left[ \left\{ \left( \frac{8}{4}+2 \right)-0 \right\}-\left( \frac{8}{3}-0 \right) \right]\] \[=2\left[ 2+2-\frac{8}{3} \right]=\frac{8}{3}\] sq. units.