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Wikipedia has the following: equation of the tangent line at a point $(a,b)$ such that $f(a,b) = 0$ (the implicit function) is given by:

${\partial f \over \partial x} (x-a) + {\partial f \over \partial y}(y-b) = 0$

I guess it's related to the implicit function theorem, which I know (that the said theorem exists, not that I can prove it myself or even be acquitted with it).

Explanation or non-rigorous outline of proof would suffice.

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    $\begingroup$ I'm at a loss why people want this closed. It's a fine question to me. $\endgroup$ – Lord_Farin Nov 2 '13 at 18:25
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We can think of the tangent line as all the vectors that are orthogonal to the normal of the curve at a given point.

If $f(x,y) =c$ is a curve in the plane then the normal vector is given by its gradient:

$$N = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right).$$

Since we want it to begin at point $(a,b)$ we want the set of points satisfying the following equation:

$$N \vert_{(a,b)} \cdot ( (x,y) - (a,b) ) = 0.$$

This equates to

$$\frac{\partial f}{\partial x}\bigg\vert_{(a,b)} (x-a) + \frac{\partial f}{\partial y} \bigg\vert_{(a,b)} (y-b)=0.$$

Let us check this agaisnt a specific example. Consider the unit circle given by $x^2 +y^2 =1$ and let's find the equation of the tangent line at point $(0,1)$. Using this we find $N = (2x,2y)$ and putting this into what we just got translates to

$$0 \cdot (x-0) + 2(y-1) = 0 = 2y-2,$$

therefore $y=1$.

Using standard calculus methods we can write $y$ as a function of $x$ in the upper part, obtaining $y = \sqrt{1-x^2}$. Differentiating and setting $x=0$ will give us the slope at that point:

$$\frac{dy}{dx} \bigg\vert_{x=0} = \frac{1}{2} \cdot \frac{-2x}{\sqrt{1-x^2}} \Bigg\vert_{x=0} = \frac{-x}{\sqrt{1-x^2}} \Bigg\vert_{x=0} = 0.$$

Using the equation of a line we get $y -1 = 0(x-0)$ and, not surprisingly, we find $y=1$.

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