### Area Under Curves Question 9

#### Question 9 - 2024 (30 Jan Shift 2)

Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y-y=Y^{\prime}(x)(X-x)$ and the co-ordinate axes, where $(x, y)$ is any point on the curve, is always $\frac{-y^{2}}{2 Y^{\prime}(x)}+1, Y^{\prime}(x) \neq 0$. If $Y(1)=1$, then $12 Y(2)$ equals

## Show Answer

#### Answer (20)

#### Solution

$A=\frac{1}{2}\left(\frac{-y}{Y^{\prime}(x)}+x\right)(y-x Y / x)=\frac{-y^{2}}{2 Y^{\prime}(x)}+1$

$\left(-y+xY^{\prime}(x)\right)\left(y-x Y^{\prime}(x)\right)=-y^{2}+2 Y^{\prime}(x)$

$-y^{2}+x y Y^{\prime}(x)+x y Y^{\prime}(x)-x^{2}\left[Y^{\prime}(x)\right]^{2}$

$2 x y-y^{2} y^{\prime}(x)=2 Y^{\prime}(x)$

$\frac{d y}{d x}=\frac{2 x y-2}{x^{2}}$

$\frac{d y}{d x}-\frac{2}{x} y=\frac{-2}{x^{2}}$

I.F. $=e^{-2 \ln x}=\frac{1}{x^{2}}$

$y \cdot \frac{1}{x^{2}}=\frac{2}{3} x^{-3}+c$

Put $x=1, y=1$

$1=\frac{2}{3}+c \Rightarrow c=\frac{1}{3}$

$Y=\frac{2}{3} \cdot \frac{1}{x}+\frac{1}{3} X^{2}$

$\Rightarrow 12 Y(2)=\frac{5}{3} \times 12=20$