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Considering $f:\Bbb R^2 \rightarrow \Bbb R$ defined by $ f(x,y) = x^2-y^2$ and $S$ the unit circle, I wish to understand why in two dimensions, the condition that $\nabla f (x_0) = \lambda \nabla g(x_0)$ is the same as the level curves being tangent at $x_0$ which is stated as part of a solution to a Lagrange Multipliers problem in my multi-variable calculus textbook.

The sketch of this problem shows the level curves of $f(x,y)$ (hyperbolas except for the level set $f=0$ consisting of the lines $y=x, y=-x$) intersecting the unit circle where only at the points $(1, 0) , (-1,0), (0,1), (0,-1)$ are the two curves tangent. Any insights appreciated.

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Fix a function $h : \Bbb R^2 \to \Bbb R$ and a point $x_0 \in \Bbb R^2$, and suppose there is a $C^1$ curve $\gamma(t) : J \to \Bbb R^2$ satisfying $\gamma(0) = x_0$ and $\gamma'(0) \neq 0$ and whose image is contained inside the level curve through $x_0$. Then $h(\gamma(t)) = h(x_0)$ for all $t \in J$, so $$(\nabla h)(x_0) \cdot \gamma'(0) = \nabla_{\gamma'(0)} h = (h \circ \gamma)'(0) = 0 , $$ and thus the tangent line to the level set of $h$ through $x_0$ is contained in $(\nabla h)(x_0)^{\perp}$. (If $(\nabla h)(x_0) \neq 0$, these coincide.)

Now, if $(\nabla f)(x_0)$ and $(\nabla g)(x_0)$ are nonzero and parallel, their spans coincide and hence so do their orthogonal lines: $$(\nabla f)(x_0)^{\perp} = (\nabla g)(x_0)^{\perp} .$$

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