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
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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 $x=1, y=1$
$1=\frac{2}{3}+c \Rightarrow c=\frac{1}{3}$
$Y=\frac{2}{3} \cdot \frac{1}{x}+\frac{1}{3} X^{2}$
$\Rightarrow 12 Y(2)=\frac{5}{3} \times 12=20$
