A) 3
B) 6
C) 9
D) None of these
Correct Answer: C
Solution :
Given parabola, is \[{{(y-2)}^{2}}=x-1\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{2(y-2)}\] when,\[y=3,\,\,x=2\] \[\therefore \] \[\frac{dy}{dx}=\frac{1}{2(3-2)}=\frac{1}{2}\] Tangent at\[(x,\,\,3)\]is \[y-3=\frac{1}{2}\,(x-2)\] \[\Rightarrow \] \[x-2y+4=0\] \[\therefore \]Required area \[\int_{0}^{3}{\{{{(y-2)}^{2}}+1\}dy-\int_{0}^{3}{(2y-4)dy}}\] \[=\left[ \frac{{{(y-2)}^{3}}}{3}+y \right]_{0}^{3}-[{{y}^{2}}-4y]_{0}^{3}\] \[=\frac{1}{3}+3+\frac{8}{3}-(9-12)=9\]You need to login to perform this action.
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