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The equation $F(x,y)=0$ defines a curve $C$ in $\mathbb{R}^2$ consisting of points $(x,y)$ which satisfy this equation. Assume $F$ is continuously differentiable. Also assume that $\nabla F$ is nonzero at every point of $C$. Let $(x_0,y_0)$ belong to $C$.

Using the above, find a formula for the tangent line $T$ at $(x_0,y_0)$ in terms of the above quantities.

My attempt: Intuitively, I want to utilize that the approximate slope is $\frac{x-x_0}{y-y_0}$, i.e. we have $m(x-x_0)=(y-y_0)$, where the gradient is somehow used to give $m$? I fear I'm oversimplifying this, any idea where I should go?

Any help appreciated!

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    What's wrong with $y-f(x_0) = f'(x_0)(x-x_0)$? – 2017-01-30
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    The gradient is orthogonal to the level curve. So, you can get the tangent line as the line orthogonal to $\nabla F$. – 2017-01-30

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