Mathematics - Area Under Curves Question with Solution | TestHub

The equation of the parabola is $y^2-4y-x+5=0$ The equation of tangent at $(2,3)$ is $$\begin{array}{rl}3y-2(y+3)-\frac{x+2}{2}+5& =0\\ 2y-x-4& =0\end{array}$$
$\therefore$ Required area $A$ is given by $$\begin{array}{rl}& A={\int }_0^3\left(x_2-x_1\right)dy\\ & \Rightarrow A={\int }_0^3\left[\left\{(y-2{)}^2+1\right\}-\{2y-4\}\right]dy\\ & \Rightarrow A={\int }_0^3\left(y^2-6y+9\right)dy={\int }_0^3(3-y{)}^{2}dy\\ & =-{\left[\frac{(3-y{)}^3}{3}\right]}_0^3=9\\ & \end{array}$$