Ellipse equation in $\Bbb R^3$

Question

I'm asked to prove that the system of equations: $x^2+y^2=1, y=z$ represents an ellipse. I understand that the shadow of this shape on the plane $xy$ is a circle. Now letting $z=y$ seems to me that it is a circle that was stretched along the $z-axis$ to form an ellipse. Other than this, I have no idea where to start.

As a hint, I get that i need to picture these equations on the coordinate system where the plane $0xy$ is the plane $y=z$. So I'm guessing this can be proved with a system change?

Answer 1

Let's look at a local to global rotation $$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{vmatrix} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{vmatrix} \begin{pmatrix} x \\ u \\ v \end{pmatrix} $$

If the curve can be described in local coordinates with $v=0$ it means it lies on a plane (as the local $v$ direction is away from the plane). Using $y=z$ translates to $u \cos\theta = v \sin \theta$ or $\theta = \frac{\pi}{4}$.

So now we have $$\begin{align} y &= \frac{u}{\sqrt{2}} \\ z &=\frac{u}{\sqrt{2}} \end{align}$$

and the given curve is

$$ \left. x^2 + \left( \frac{u}{\sqrt{2}} \right)^2 = 1 \right\} x^2 + \frac{1}{2} u^2 = 1 $$

which is an ellipse on the plane with semi radii $1$ and $\sqrt{2}$.

Answer 2

HINT:

Take a cylinder radius $1$ and cut it by plane $y=z$ and consider the intersection.Plane $y-z$ is slanted at $45^0$ to planes $Oyx$ and $Oyz.$

Answer 3

This is essentially the same answer as ja72's but from a slightly different perspective.

Let $S= \{x | x_1^2+x_2^2 = r, x_2=x_3 \} = \{x | f(x) = 0, x_2=x_3 \}$, where $f(x) = x_1^2+x_2^2 - r$.

Since $x_2=x_3$, we see that $S$ lies in the subspace $\operatorname{sp} \{ (0,1,-1)^T \}^\bot$, so we choose an orthogonal basis $b_1 = {1 \over \sqrt{2}} (0,1,-1)^T, b_2 = {1 \over \sqrt{2}} (0,1,1)^T, b_3(1,0,0)^T $ and let $B=[b_1\,b_2\,b_3]$.

Now we have \begin{eqnarray} S' &=& B^{-1} S \\ &=& \{B^{-1} x | f(x) = 0, x_2=x_3 \} \\ &=& \{ y | f(By) = 0, y_1 = 0 \} \\ &=& \{ (0,a,b)^T | f(B(0,a,b)^T = 0 \} \\ &=& \{ (0,a,b)^T | f((b, {1 \over \sqrt{2}}a, {1 \over \sqrt{2}}a)^T) \} \\ &=& \{ (0,a,b)^T | b^2 + {a^2 \over 2} = r \} \end{eqnarray} Which is seen to be an ellipse.