Let Y=Y(X) be a curve lying in the first quad | vedclass

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(A) The equation of the tangent at $(x, y)$ is $Y-y=Y^{\prime}(x)(X-x)$.For $X=0$,$Y=y-x Y^{\prime}(x)$.For $Y=0$,$X=x-\frac{y}{Y^{\prime}(x)}$.The ar

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$20$

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(A) The equation of the tangent at $(x, y)$ is $Y-y=Y^{\prime}(x)(X-x)$.For $X=0$,$Y=y-x Y^{\prime}(x)$.For $Y=0$,$X=x-\frac{y}{Y^{\prime}(x)}$.The area of the triangle formed by the tangent and the coordinate axes is $A = \frac{1}{2} \left| x - \frac{y}{Y^{\prime}(x)} ight| \left| y - x Y^{\prime}(x) ight|$.Since the curve is in the first quadrant and the area is given as $\frac{-y^2}{2 Y^{\prime}(x)}+1$,we have:$A = \frac{1}{2} \left( \frac{x Y^{\prime}(x) - y}{Y^{\prime}(x)} ight) (y - x Y^{\prime}(x)) = \frac{-y^2}{2 Y^{\prime}(x)} + 1$.Multiplying by $2 Y^{\prime}(x)$:$-(y - x Y^{\prime}(x))^2 = -y^2 + 2 Y^{\prime}(x)$.$-y^2 + 2xy Y^{\prime}(x) - x^2 (Y^{\prime}(x))^2 = -y^2 + 2 Y^{\prime}(x)$.$2xy Y^{\prime}(x) - x^2 (Y^{\prime}(x))^2 = 2 Y^{\prime}(x)$.Since $Y^{\prime}(x) 
eq 0$,we divide by $Y^{\prime}(x)$:$2xy - x^2 Y^{\prime}(x) = 2$.$Y^{\prime}(x) = \frac{2xy - 2}{x^2} = \frac{2y}{x} - \frac{2}{x^2}$.This is a linear differential equation: $\frac{dy}{dx} - \frac{2}{x} y = -\frac{2}{x^2}$.Integrating factor $I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$.The solution is $y \cdot \frac{1}{x^2} = \int \left( -\frac{2}{x^2} \cdot \frac{1}{x^2} ight) dx = \int -2 x^{-4} dx = \frac{2}{3} x^{-3} + C$.$y = \frac{2}{3x} + C x^2$.Given $Y(1)=1$,$1 = \frac{2}{3} + C \Rightarrow C = \frac{1}{3}$.So,$Y(x) = \frac{2}{3x} + \frac{x^2}{3}$.$Y(2) = \frac{2}{3(2)} + \frac{4}{3} = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$.$12 Y(2) = 12 	imes \frac{5}{3} = 20$.

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

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