# Find the point(s) (if any) of horizontal tangent lines for the equation x^2+xy+y^2=6. If none exist, why?

Nov 11, 2016

graph{(x^2+xy+y^2-6)((x-sqrt2)^2+(y+2sqrt2)^2-0.01)((x+sqrt2)^2+(y-2sqrt2)^2-0.01)=0 [-8.89, 8.89, -4.444, 4.445]}

${P}_{1} = \left(\sqrt{2} , - 2 \sqrt{2}\right)$
${P}_{2} = \left(- \sqrt{2} , + 2 \sqrt{2}\right)$

#### Explanation:

Implicitly differentiating over x the equation we get (using y'=$\frac{\mathrm{dy}}{\mathrm{dx}}$)

$2 x + y + x y ' + 2 y y ' = 0$

So

$2 x + y = - \left(x + 2 y\right) y '$

$y ' = - \frac{2 x + y}{x + 2 y}$

So $y ' = 0$ only when $y = - 2 x$ (excluding $y = 0$ when $y ' = - 2$)

Now intersect with the equation

$\left\{\begin{matrix}y = - 2 x \\ {x}^{2} + x y + {y}^{2} = 6\end{matrix}\right.$
$\left\{\begin{matrix}y = - 2 x \\ {x}^{2} - 2 {x}^{2} + 4 {x}^{2} = 6\end{matrix}\right.$
$\left\{\begin{matrix}y = - 2 x \\ 3 {x}^{2} = 6\end{matrix}\right.$
$\left\{\begin{matrix}y = - 2 x \\ x = \pm \sqrt{2}\end{matrix}\right.$