A) 3
B) 6
C) 9
D) 12
Correct Answer: C
Solution :
Given equation of parabola is\[{{y}^{2}}-4y-x+5=0\] Then, the equation of tangent at (2, 3) is \[3y-2(y+3)-\frac{(x+2)}{2}+5=0\]\[\Rightarrow \]\[2y-x-4=0\] \[\therefore \]Required area is given by \[A=\int_{0}^{3}{({{x}_{2}}-{{x}_{1}})dy}\] \[=\int_{0}^{3}{[{{\{y-2)}^{2}}+1\}-\{2y-4\}]}dy\] \[=\int_{0}^{3}{({{y}^{2}}-6y+9)}d=\int_{0}^{3}{{{(3-y)}^{2}}}dy\] \[=-\left[ \frac{{{(3-y)}^{3}}}{3} \right]_{0}^{3}\]\[\therefore \]A = 9You need to login to perform this action.
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