Let slope of the tangent line to a curve at any point P(x,y) be given by xy2+yx . If the curve intersects the line x+2y=4 at x=-2, then the value of y, for which the point (3,y) lies on the curve, is :

dydx=xy2+yx xdy-ydx=xy2dxydx-xdyy2=-xdx⇒∫dxy=fxdx⇒xy=-x22+c↓ passes through (-2,3)⇒-23=-2+c⇒c=43 Curve is xy=-x22+43(3,y) lies on it ⇒3y=-92+43⇒3y=-196⇒y=-1819

Let slope of the tangent line to a curve at any point P(x,y) be given by xy2+yx . If the curve intersects the line x+2y=4 at x=-2, then the value of y, for which the point (3,y) lies on the curve, is :

dydx=xy2+yx xdy-ydx=xy2dxydx-xdyy2=-xdx⇒∫dxy=fxdx⇒xy=-x22+c↓ passes through (-2,3)⇒-23=-2+c⇒c=43 Curve is xy=-x22+43(3,y) lies on it ⇒3y=-92+43⇒3y=-196⇒y=-1819

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$Let slope of the tangent line to a curve at any point P (x, y) be given by \frac{x y^{2} + y}{x} . If the$
$curve intersects the line x + 2 y = 4 at x = - 2, then the value of y, for which the point$
$(3, y) lies on the curve, is :$

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$- \frac{4}{3}$

$(3, y) lies on the curve, is :$

$\frac{18}{35}$

$- \frac{18}{19}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{d y}{d x} = \frac{x y^{2} + y}{x} x d y - y d x = x y^{2} d x \\ \frac{y d x - x d y}{y^{2}} = - x d x \Rightarrow \int d (\frac{x}{y}) = f x d x \Rightarrow \frac{x}{y} = \frac{- x^{2}}{2} + c \\ ↓ passes through (- 2, 3) \Rightarrow \frac{- 2}{3} = - 2 + c \Rightarrow c = \frac{4}{3} \\ Curve is \frac{x}{y} = \frac{- x^{2}}{2} + \frac{4}{3} \\ (3, y) lies on it \Rightarrow \frac{3}{y} = \frac{- 9}{2} + \frac{4}{3} \Rightarrow \frac{3}{y} = \frac{- 19}{6} \Rightarrow y = \frac{- 18}{19} \end{array}$