How does this equation $x^2+y^2+xy=1/2$ represent an ellipse?

Question

I want to find the intersection of the sphere $x^2+y^2+z^2 = 1$ and the plane $x+y+z=0$.

$z=-(x+y)$ that gives $x^2+y^2+xy= \frac 12$

How do I represent this in the standard form of ellipse? Any help is appreciated to proceed further. Thanks in advance.

Answer 1

$$x^2+y^2+xy=1/2 \implies \frac{(x+y)^2+(x-y)^2}{2}+\frac{(x+y)-(x-y)^2}{4}=\frac{1}{2}$$ $$\implies \frac{3}{2}(x+y)^2+\frac{(x-y)^2}{2}=1$$ $$\implies \frac{(\frac{x+y}{\sqrt{2}})^2}{1/3}+\frac{(\frac{x-y}{\sqrt{2}})^2}{1}=1,$$ which is an ellipse.

Answer 2

$$\dfrac12=\left(x+\dfrac y2\right)^2+y^2\left(1-\dfrac14\right)$$

Let $x+\dfrac y2=X, y=Y$

Alternately, use this to eliminate $xy$

Also, from the link

(the coefficient of $xy)^2-4\cdot$(the coefficient of $x^2)\cdot($(the coefficient of $y^2)$

$$=(1)^2-4\cdot1\cdot1<0$$

Hence it's an ellipse

Answer 3

Let $\sqrt{2}x = u+v$ and $\sqrt{2}y=u-v$. Then the resulting expression is

$$3u^2 + v^2 = 1$$

which is the standard form of an ellipse and since the transformation is a pure rotation, the shapes of the objects haven't been distorted at all.

Answer 4

As this quadrar=tic of $x$ and $y$ has $xy$ term, it cannot represent a circle. $$x^2+y^2-xy=1/2~~~(1)$$ Next, write it as quadratic of $y$: $$y^2+xy+x^2-1/2=01/2 \implies y=\frac{-x\pm\sqrt{x^2-4x^2+2}}{2}$$. For the curve to be real $$3-3x^2 \ge 0 \implies -\sqrt{2/3}\le x \le \sqrt{2/3}$$ Therefore it is a bounded conic it has to be an ellipse as a circle was ruled out.