# Level curve tangents

My math course comes back to multivariate differentiations and dives right into it without a proper refresher. could someone help me on the road on how to crack this problem?

Close to the point (2,0), the curve $x^2+xy+y^2=4$ defines y as an implicit function of x.

Which points on the curve have a horizontal tangent?

• The curve has a horizontal tangent at the points where the implicit function $y(x)$ has derivative $0$. To find those points, implicitly differentiate and solve. Mar 12, 2018 at 16:01
• is there a way to express $y(x)$ that I oversee? because I don't see how to get y out without using the quadratic formula
– SOMI
Mar 12, 2018 at 16:11
• You're not supposed to. That's what implicit differentiation is all about. Whatever function $y(x)$ is, we know that it fulfills $x^2 + x\cdot y(x) + (y(x))^2 = 4$. Since this equality is an equality of functions of $x$ (and not an equation or something), we can differentiate with respect to $x$ on both sides. Then you set $y'(x) = 0$ and you get two equations with two unknowns that you can solve. Mar 12, 2018 at 16:18
• Can someone provide me a complete solution?
– SOMI
Mar 12, 2018 at 16:29

We have that

$$x^2+xy+y^2=4 \implies 2xdx+ydx+xdy+2ydy=0\implies(x+2y)dy=-(2x+y)dx \implies \frac{dy}{dx}=-\frac{2x+y}{x+2y}=0 \implies y=-2x$$

thus

$$x^2+x(-2x)+(-2x)^2=4\implies 3x^2=4 \implies x=\pm\frac{2\sqrt 3}{3} \quad y=\mp\frac{4\sqrt 3}{3}$$

As an alternative find the explicit form

$$y(x)=\frac{-x\pm\sqrt{16-3x^2}}{2}$$

and evaluate $y'(x)=0$.