The slope of the tangent to a curve at P(x,y) is xy^2+yx.

Question

The slope of the tangent to a curve at $P(x,y)$ is $\dfrac{xy^2+y}{x}$. If the curve intersects $x+2y=4$ at $x=-2$, the value of $y$ when the point $(3,y)$ lies on the curve is:

Accepted Answer

$$\frac{xdy-ydx}{y^2}=x\,dx \implies -\frac{x}{y}=\frac{x^2}{2}+C.$$

Curve passes through $(-2,3)$ (since $x=-2$ on $x+2y=4$ gives $y=3$): $\dfrac{2}{3}=2+C \Rightarrow C=-\dfrac43$.

At $x=3$: $-\dfrac3y = \dfrac92-\dfrac43 \Rightarrow y=-\dfrac{18}{19}$.

Solutions · 1