2
$\begingroup$

I want to know what is the intersection of the plane $2x+y+z-2=0$ and $x^2+y^2=z^2$. My guess is that it is an ellipse or an hyperbola, but I can't understand how to determine which one.

Subsituting $z=2-2x-y$ in the cone equation, we obtain $$ 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0 $$ and I can't undertsand whether it is an hyperbola or an ellipse.

Will you please help me figure it out?

Thanks

$\endgroup$
2
  • $\begingroup$ How does the quadratic part $5x^2+4xy+2y^2$ factorise? $\endgroup$ Jun 24 '18 at 8:15
  • $\begingroup$ Compare with $ax^2+2hxy+by^2+...=0$ and find $ab-h^2$. It its zero then its a parabola, if its positive then its ellipse else its hyperbola $\endgroup$ Jun 24 '18 at 8:15
1
$\begingroup$

You have $$ \tag{*} 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0 $$ and your goal is to figure out whether that is an ellipse or a hyperbola. We can rephrase that question as, "is it possible that the equation can keep having solutions for large coordinate values?"

For large sizes of the coordinates, the quadratic terms $$ 5x^2 + 4xy + 2y^2 $$ will dominate, but it is not immediately clear whether they can keep canceling out each other, because the sign of $4xy$ can be both positive and negative. But we can get rid of the mixed term by completing the square together with one of the other terms -- for example, $$ 2(x+y)^2 = 2x^2 + 4xy + 2y^2 $$ so $$ 5x^2 + 4xy + 2y^2 = 5x^2 + 2(y+x)^2 - 2x^2 = 3x^2 + 2(x+y)^2 $$ which is large and positive for large coordinate values and therefore cannot be canceled out by the linear terms which grow slower.

So this shape must be an ellipse (or a point, or nothing at all).


Unfortunately, though, your equation (*) is not right.

On geometric grounds, if the intersection is an ellipse, then $(0,0)$, which is on the axis of the cone, ought to be inside the ellipse. However, the LHS of (*) is positive at $(0,0)$, and we have just found that it is also positive when the coordinates are large. That doesn't make sense. So something must be wrong.

It looks like you inserted your $z=2-2x-y$ into $x^2+y^2+z^2=0$ instead of into $x^2+y^2=z^2$. So now you need to go back and do something like the above for the equation you get after fixing this sign error.

$\endgroup$
1
$\begingroup$

Making the change of variables ( a rigid rotation does not alter the form)

$$ x = X\cos\theta-Y\sin\theta\\ y = X\sin\theta+Y\cos\theta $$

in

$$ f(x,y) = 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0 $$

we get at

$$ f(x,y)\equiv F(X,Y) = \frac{1}{2} (4 \sin (2 \theta)+3 \cos (2 \theta)+7)X^2+(4 \cos (2 \theta)-3 \sin (2 \theta))X Y+\frac{1}{2} (7-4 \sin (2 \theta)-3 \cos (2 \theta))Y^2-12\cos\theta X+12\sin\theta Y+4=0 $$

and now choosing $\theta$ such that $4\cos(2\theta)-3\sin(2\theta) = 0\to \theta = \frac 12 \arctan(\frac 43)$ we obtain finally

$$ F(X,Y) =6 X^2-\frac{24 X}{\sqrt{5}}+Y^2+\frac{12 Y}{\sqrt{5}}+4 = 0 $$

which is clearly an ellipse.

NOTE

As can be verified the intersection of the cone $x^2+y^2=z^2$ and the plane $2x+y+z-2=0$ gives

$$ f(x,y) = 4 - 8 x + 3 x^2 - 4 y + 4 x y = 0 $$

and not

$$ f(x,y) = 5 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0 $$

as assumed. In this case with the same rotation the true intersection curve should be

$$ F(X,Y) = 4 X^2-4 \sqrt{5} X-Y^2+4 $$

which is a hyperbola.

$\endgroup$
1
  • $\begingroup$ A wrong sign change was responsible for the modifications in the answer. Sorry. $\endgroup$
    – Cesareo
    Jun 24 '18 at 9:31
0
$\begingroup$

There was probably a typo error or calculation error in coefficient of $x^2$. It should be

$$ 3 x^2 + 4 x y - 8 x + 2 y^2 - 4 y + 4 = 0 $$

which as others also pointed out, by considering discriminant $ ab -h^2 <0 $ should be a hyperbola.

$\endgroup$
0
$\begingroup$

Let me offer a more geometrical approach. To check if the intersection is an ellipse, a parabola or a hyperbola it is enough to check whether the plane intersects all the generatrices of the cone or not. In practice, this can be done as follows.

  1. Construct the vector $\vec n$ perpendicular to the plane; in your case you can read it off the equation of the plane: $\vec n=(2,1,1)$.

  2. Construct a vector $\vec t$ perpendicular to $\vec n$ and such that plane $(\vec n,\vec t)$ is parallel to the axis of the cone. If $\vec l$ is any vector along the axis, then a suitable choice is $$\displaystyle\vec t={\vec n\cdot\vec n\over\vec n\cdot\vec l}\ \vec l-\vec n.$$ In your case you can take $\vec l=(0,0,1)$ to get $\vec t=(-2,-1,5)$.

  3. Check if point $P=V+\vec t$ is inside the cone or not, where $V$ is the cone vertex. In your case $V=(0,0,0)$ and $P=(-2,-1,5)$; if $P$ is inside the cone its coordinates should satisfy $x^2+y^2<z^2$, as it is indeed the case.

  4. If $P$ is inside the cone, then the intersection is a hyperbola (your case); if $P$ lies on the cone surface, then the intersection is a parabola; if $P$ is outside the cone, then the intersection is an ellipse.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.