A) \[\frac{17}{3}sq\,unit\]
B) \[\frac{19}{3}sq\,unit\]
C) \[9\,sq\,unit\]
D) \[15\,sq\,unit\]
Correct Answer: C
Solution :
The point of intersection of \[{{y}^{2}}=4x\] and \[y=2x-4\]is \[{{(2x-4)}^{2}}=4x\] \[\Rightarrow \] \[{{x}^{2}}-5x+4=0\] \[\Rightarrow \] \[(x-1)(x-4)=0\] \[\Rightarrow \] \[x=1,4\] \[\Rightarrow \] \[y=-2,4\] \[\therefore \] Required area \[=\int_{-2}^{4}{\left( \frac{y+4}{2} \right)dy-\int_{-2}^{4}{\frac{{{y}^{2}}}{4}}dy}\] \[=\frac{1}{2}\left[ \frac{{{y}^{2}}}{2}+4y \right]_{-2}^{4}-\frac{1}{4}\left[ \frac{{{y}^{3}}}{3} \right]_{-2}^{4}\] \[=\frac{1}{2}[8+16-(2-8)]-\frac{1}{12}[64+8]\] \[=\frac{1}{2}[30]-\frac{1}{12}(72)\] \[=15-6=9sq\,unit\]You need to login to perform this action.
You will be redirected in
3 sec