I have the curve $x^2+xy+y^2=1$, whose graph is as follows:
I'm trying to find the points where the line tangent to the curve is parallel to the line $y = -x$. There are colleagues of mine who answered the points (0.5; 0.5) and (-0.5; -0.5). I tried to solve by doing the implicit derivation and equating it to -1, which is the slope of the tangent line:
$$\frac{d}{dx}(x^2+xy+y^2)=\frac{d}{dx}(1)\leftrightarrow 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\leftrightarrow \frac{dy}{dx}=\frac{-2x-y}{x+2y}\leftrightarrow \frac{dy}{dx}=-\frac{2x+y}{x+2y}=-1\leftrightarrow \frac{2x+y}{x+2y}=1\leftrightarrow 2x+y=x+2y\leftrightarrow y=x$$
The problem is that I have two variables and an equation. I already thought of replacing $y$ with $x$ in the initial curve, but that doesn't respect the condition of the tangent line. How can I find the values of $x$ and $y$, given that I only have one equation?
All the parallels to $y + x = 0$ have the representation $y+x=m$
Substituting $y = m-x$ into the ellipse equation we have
$$ x^2 + x(m-x)+(x+m)^2 = 1 $$
now solving for $x$
$$ x = \frac 12\left(m\pm \sqrt{4-3m^2}\right) $$
but at tangency we should have $4-3m^2=0$ so the tangent lines are
$$ y+x=\pm\sqrt{\frac 43} $$
at
$$ x =\frac 12 \left\{-\sqrt{\frac 43}, \sqrt{\frac 43}\right\} $$