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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.

Aniruddha Deb
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Joyce
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  • Note: Substituting $z=-(x+y)$ won't give you the intersection of the sphere and the plane. It gives you the shadow of that intersection in the $xy$-plane. The actual intersection is a circle in space. Is that what you want to find? – Blue Apr 08 '20 at 05:45
  • There are formulae for getting the standard form, including the center and angle of rotation on the Wikipedia page. It looks like it's centered at the origin and rotated by $\pi/4$. I'll leave $a,b$ to you. –  Apr 08 '20 at 05:46
  • @Blue I wanna find the projection onto xy plane, which is an ellipse? – Joyce Apr 08 '20 at 05:51
  • @Stackuser: The projection is definitely an ellipse, as the various answers show. (It also makes sense when you think about how the shadow of a circle should appear.) – Blue Apr 08 '20 at 06:02

4 Answers4

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

Z Ahmed
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  • I would like to know the reason behind from step 2 to step 3. Can $a$ be $2/3$ and $b$ equals 2 from step 2? Could you please let me know what I am missing here? – Joyce Apr 08 '20 at 05:55
  • $$Stackuser Good questio, Every ellipse can be written as $L_1^2/a^2+L_2^2/b^2=1$ If $L_1,L_2$ nutually perpendicular and normalised then only$a$ and $b$ are semi-mejor aor semi-minor ases. By normalized, one means the equation of a line as $\frac{px+qx+r}{\sqrt{p^2+q^2}}=0$. the lines$ x=0, y=0, x=2, y=3$ are self normalized, In this example $a=1/\sqrt{3},, b=1$. – Z Ahmed Apr 08 '20 at 06:23
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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.

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

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

lab bhattacharjee
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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.

Z Ahmed
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