<s>[INST] How to find the Minimum of the Function. Minimize $xy$ on the ellipse $b^2x^2+a^2y^2=a^2b^2$
So what I did first was take the gradient of $f$ and $g$ 
$∇ f = (y,x)$    $\qquad$            $∇ g= (2x^2b^2,a^22y)$ 
Then we do the lagrange multiplier 
$y= 2x^2b^2λ$
$x=2ya^2λ$
Then we equate the functions. By multiplying.
$(y= 2x^2b^2λ)y^2a^2$
$(x=2ya^2λ )x^2b^2$

$y^3=2x^2y^2a^2b^2λ$
$x^3=2x^2ya^2b^2λ$
But this is where I get stuck. Setting them equal to each other gets me this.
$2x^2y^2a^2b^2= 2x^2ya^2b^2$
$y=1$ 
I think there is another method to solve this problem than the one I specifed does anyone know what that method is?
 [/INST] $$L(x,y,z)=xy+\lambda(b^2x^2+a^2y^2-a^2b^2)$$
$$\nabla_{x,y,\lambda} L(x,y,z)=\langle y+2\lambda b^2 x, x+2\lambda a^2 y , b^2x^2+a^2y^2-a^2b^2 \rangle=\langle 0,0,0 \rangle$$

$$y=-2 \lambda b^2 x$$
$$x=-2 \lambda a^2 y$$

assuming $\lambda \ne0$:
$$\frac{y}{x}=-2 \lambda b^2$$
$$\frac{y}{x}=\frac{1}{-2 \lambda a^2}$$

$$-2 \lambda b^2=\frac{1}{-2 \lambda a^2}$$
or 
$$\lambda=\frac{\pm1}{2 a b}$$
$$y=\mp \frac{ b}{a}x$$
$$b^2x^2+a^2(\frac{b^2}{a^2})x^2=a^2b^2$$
$$x=\pm\frac{a}{\sqrt2}$$
$$y=\mp\frac{b}{\sqrt2}$$

$$xy=\frac{-ab}{2}$$
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