Find rectangular Cartesian coordinate system

Question

Find rectangular Cartesian coordinate system in $\mathbb{R}^3$ to bring quadratic surface $2y^2-3z^2+4xz-12y+15=0 $ to standard form.

After splittig off square the equation can be rewritten as follows $2(y-3)^2-3z^2+4xz=3$.

What to do with mixed term $4xz$?

Answer 1

Just because it's kind of fun, here's a method using the Matrix Representations of Conic Sections. Essentially, for $$ Q(x,z) = -3z^2 + 4xz$$ We can write \begin{align} Q(x,z) &= \begin{pmatrix} x & z \end{pmatrix} \begin{pmatrix} 0 & 2\\ 2 & -3 \end{pmatrix} \begin{pmatrix} x \\ z \end{pmatrix} \end{align} We can obtain the linear transformation that eliminates the cross-term $4xz$ by diagonalizing the above matrix (or in general, the best we can do is the Jordan Normal Form).

In general, the normal form for a matrix $M$ looks like $$ M = S J S^{-1}$$ where $J$ is a block-diagonal matrix (in this case, hopefully a completely diagonal matrix).

Wolfram Alpha leads to \begin{align} \underbrace{\begin{pmatrix} 0 & 2\\ 2 & -3 \end{pmatrix}}_{M} &= \underbrace{ \begin{pmatrix} -1 & 2\\ 2 & 1 \end{pmatrix}}_{S} \underbrace{ \begin{pmatrix} -4 & 0\\ 0 & 1 \end{pmatrix}}_{J} \underbrace{ \begin{pmatrix} -1 & 2\\ 2 & 1 \end{pmatrix}^{-1}}_{S^{-1}}\\ &= \begin{pmatrix} -1 & 2\\ 2 & 1 \end{pmatrix} \begin{pmatrix} -4 & 0\\ 0 & 1 \end{pmatrix} \begin{pmatrix} -1/5 & 2/5 \\ 2/5 & 1/5 \end{pmatrix} \end{align}

If we use the coordinate transformation \begin{align} \begin{pmatrix} x' \\ z' \end{pmatrix} &= S^{-1} \begin{pmatrix} x \\ z \end{pmatrix}\\ &= \begin{pmatrix} -1/5 & 2/5 \\ 2/5 & 1/5 \end{pmatrix} \begin{pmatrix} x \\ z \end{pmatrix}\\ &= \begin{pmatrix} (2x - z)/5\\ (2x + z)/5 \end{pmatrix} \end{align}

To see this in action, notice also that \begin{align} \begin{pmatrix} x \\ z \end{pmatrix} &= S \begin{pmatrix} x' \\ z' \end{pmatrix}\\ &= \begin{pmatrix} -1 & 2\\ 2 & 1 \end{pmatrix} \begin{pmatrix} x' \\ z' \end{pmatrix}\\ &= \begin{pmatrix} 2z' - x'\\ z' + 2x' \end{pmatrix} \end{align}

If we just let $y' = y - 3 \iff y = y' + 3$ as you've demonstrated, then subbing into the quadratic surface cartesian equation that you have gives, \begin{align} 2y^2 - 3z^2 + 4xz - 12y + 15 &= 0\\ \Rightarrow 2(y' + 3)^2 - 3(z' + 2x')^2 + 4(2z' - x')(z' + 2x') - 12(y' + 3) + 15 &= 0\\ \Rightarrow -20x'^2 + 2y'^2 + 5z'^2 - 3 &= 0 \end{align} Where we've used Wolfram Alpha to simplify the equation for us.

EDIT: Here are some fun pictures to along with this.

If we set $y = 3$ to ignore the $y$ dependence for a moment, we essentially have rotated from this hyperbola

to this hyperbola

by using the change of coordinates given by lines \begin{align} x &= (2x - z)/5\\ z &= (2x + z)/5 \end{align}

Answer 2

Let $A=\begin{bmatrix} 0 & 0 & 2\\ 0 & 2 & 0\\ 2 & 0 & -3\\ \end{bmatrix}$ and $X=[x\;y\;z]^T$ and your equation can be written as:
$X^TAX+[0\; -12\; \;\;\;0]X=0$.
Since $A$ is symmetric, it can be diagnolized. You may verify that eigenvalues of $A$ are $1,2,-4$ (form a diagonal matrix $D$ with these eigenvalues on the diagonal), find the corresponding orthonormal eigenvectors (these exist because matrix is symmetric and by use of Gram Schmidt process) and form an orthogonal matrix $Q$ such that $AQ= QD\implies D=Q^TAQ$
Thus,
$D=\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & -4\\ \end{bmatrix}$ , $Q=\begin{bmatrix} 2/\sqrt{5} & 0 & -1/\sqrt{5}\\ 0 & 1 & 0\\ 1/\sqrt{5} & 0 & 2/\sqrt{5}\\ \end{bmatrix}$ , $Q^TQ=QQ^T=I$, identity matrix.

Now let $X=QY=Q[x'\;\;y'\;\;z']^T$ so that the given equation becomes: $Y^T(Q^TAQ)Y+[0\; -12\; \;\;\;0][x'\;\;2y'\;\;-4z']^T=0$ $\implies Y^TDY-24y'=0 \implies x'^2+2y'^2-4z'^2-24y'=0 $. Thus we have removed the cross term.
Can you take it from here?

Answer 3

Continue with $2(y-3)^2-3z^2+4xz=3$ and, to decouple the $xz$-term, determine the conic axes in the $xz$-plane with $f(x,z)= 4xz-3z^2$,

$$\frac{f_x’}{f_y’}=\frac{4z}{4x-6z}= \frac xz\implies (z+2x)(2z-x)=0$$ which shows that the axis $z= \frac12x$ of the conic is at $\theta$-angle with respect to the $x$-axis, with $\cos\theta =\frac2{\sqrt5}$ and $\sin\theta =\frac1{\sqrt5}$. Then, the equation can be written in the standard form as

$$\left(\frac2{\sqrt5} x+ \frac1{\sqrt5}z\right)^2 + 2(y-3)^2-4\left(\frac1{\sqrt5} x-\frac2{\sqrt5}z\right)^2=3$$