Given dxdy=xxy2+y
y2xdy−ydx=xdx
−d(yx)=xdx
−yx=2x2+c
∵ Curve intersects the line x+2y=4 at x=−2⇒ point of intersection is (−2,3).
∴ Curve passes through (−2,3).
⇒32=2+c⇒c=−34
⇒y−x=2x2−34
Now put (3,y) we get,
⇒y−3=619
⇒y=19−18
Let slope of the tangent line to a curve at any point P(x,y) be given by xxy2+y. 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 :
Held on 26 Feb 2021 · Verified 6 Jul 2026.
−34
3518
−1918
−1118
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