What type of Quadric is this ? (2)

The orthogonal matrix $P \; = \; \left( \begin{array}{ccc} \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{14}} & \frac{1}{\sqrt{35}} \\ -\frac{1}{\sqrt{10}} & \frac{3}{\sqrt{14}} & \frac{3}{\sqrt{35}} \\ 0 & \frac{2}{\sqrt{14}} & -\frac{5}{\sqrt{35}}\end{array}\right)$ is such that $P^TAP \; = \; \left(\begin{array}{ccc} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 0 \end{array}\right)$ and hence the equation of the quadric becomes $35X^2 + 35Y^2 + \alpha X + \beta Y + \gamma Z + c \; = \; 0$ where $\left(\begin{array}{c} X \\ Y \\ Z \end{array}\right) \; = \; P^T \mathbf{r} \hspace{2cm} \left(\begin{array}{c} \alpha \\ \beta \\ \gamma \end{array}\right) \; = \; P^T\mathbf{b}$ We calculate that $\alpha =-7\sqrt{10}$ , $\beta = -135\sqrt{14}$ , $\gamma = 0$ , and so the equation of the quadric is $\begin{aligned} 35X^2 + 35Y^2 - 7\sqrt{10}X - 135\sqrt{14}Y + 1826 & = \; 0 \\ 35\big(X - \tfrac{1}{\sqrt{10}}\big)^2 + 35\big(Y - \tfrac{27}{\sqrt{14}}\big)^2 & = \; 0 \end{aligned}$ and so the quadric is the straight line determined by the formulae $X = \tfrac{1}{\sqrt{10}}$ , $Y = \tfrac{27}{\sqrt{14}}$ , or the vector equation $\mathbf{r} \; = \; P\left(\begin{array}{c} \frac{1}{\sqrt{10}} \\ \frac{27}{\sqrt{14}} \\ \lambda \end{array}\right) \hspace{2cm} \lambda \in \mathbb{R}$

Because of the way that Wolfram Mathematica works, some of the transpose operations had to be dropped in my solution.

$r=\{x,y,z\}, a=\left( \begin{array}{ccc} 34 & -3 & 5 \\ -3 & 26 & 15 \\ 5 & 15 & 10 \\ \end{array} \right), b=\{-156,-398,-270\}, c=1826$

$\text{Solve}[r.a.r+b.r+c=0,z]\Rightarrow \\ \left\{\left\{z\to \frac{1}{10} \left(-\sqrt{35} \sqrt{-9 x^2+6 x y+6 x-y^2-2 y-1}-5 x-15 y+135\right)\right\}, \\ \left\{z\to \frac{1}{10} \left(\sqrt{35} \sqrt{-9 x^2+6 x y+6 x-y^2-2 y-1}-5 x-15 y+135\right)\right\}\right\}$

$\text{Solve}\left[-9 x^2+6 x y+6 x-y^2-2 y-1\geq 0\right]\Rightarrow \{y\to \text{ConditionalExpression}[3 x-1,x\in \mathbb{R}]$

Fully simplified, with $3x-1$ substituted for $y$ in the solution for the original equation (above), yields $\{-5 (x-3),-5 (x-3)\}$ .

$\text{ParametricPlot3D}[\{x,3 x-1,-5 (x-3)\},\{x,-20,20\},\text{BoxRatios}\to \{1,1,1\},\text{AxesOrigin}\to \{0,0,0\}]$

Evaluating $\vec{r}^TA\vec{r}+\vec{b}^T\vec{r}+c$ gives the quadric equation: $34x^2+26y^2+10z^2-6xy+10xz+30yz-156x-398y-270z+1826=0$

Setting $y=z=0$ , I found no intersections with the x-axis.

Also, no intersection with y-axis or z-axis.

For the plane intersections below I used Excel to get it in standard form, performing a rotation and a translation of the coordinate systems. And I used it to convert solution coordinates back.

Intersection with yz-plane (x=0) is a single point: $(0,-1,15)$ .

Intersection with xz-plane (y=0) is a single point: $(\frac{1}{3},0,\frac{40}{3})$ .

Intersection with xy-plane (z=0) is a single point: $(3,8,0)$ .

Intersection with plane x=1 is a single point: $(1,2,10)$ .

Intersection with plane y=1 is a single point:: $(\frac{2}{3},1,\frac{35}{3})$ .

For all planes I chose, I got a single point, and all points found were on the line parametrized by $(x,y,z)=(3-t,8-3t,5t)$

As a check, putting these values for x, y, and z in the equation above gives $34(3-t)^2+26(8-3t)^2+10(5t)^2-6(3-t)(8-3t)+10(3-t)5t+30(8-3t)5t-156(3-t)-398(8-3t)-270(5t)+1826=0$ which is just the identity.