I have been trying this for a while. But I couldn't get it. How can I determine the point of intersection of the tangent line at (0, 0) on the curve $y^2 + y = x^3 + x^2$ ?
1 Answer
The tangent line at point $P(x_1,y_1)$ is:
$\frac{\partial f}{\partial x}(x-x_1)+\frac{\partial f}{\partial y}(y-y_1)$
Solving differentials, we obtain:
$\frac{\partial f}{\partial x}=3x^2+2x$
$\frac{\partial f}{\partial y}=-2y-1$
For $x_1=0$ and $y_1=0$ we have:
$\frac{\partial f}{\partial x}=3x^2+2x = 0$
$\frac{\partial f}{\partial y}=-2y-1 = -1$
So, by the first equation, the tangent line at $P(0,0)$ is $-y$.
Now, to calculate the intersection point with the curve we need to solve the next equations system:
$-y=0$
$y^2 + y = x^3 + x^2$
That have two solutions: $x=-1, y=0$ and $x=0, y=0$
So, the tangent line intersect the curve at $(-1,0)$ and $(0,0)$.
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3$\begingroup$ Welcome to crypto! Note: you can use $ signs inline as well to format your formulas etc. $\endgroup$– Maarten Bodewes ♦Jul 27, 2014 at 22:29