Draw the curve $8x^2+6xy-\frac x{\sqrt{10}}+3\frac y{\sqrt{10}}=1$

Question

Draw the curve Draw the curve $8x^2+6xy-\frac x{\sqrt{10}}+3\frac y{\sqrt{10}}=1$.

let $v = (x,y)\in \mathbb{R}^2$ and $B=\{(\frac 3{\sqrt{10}},\frac 1{\sqrt{10}}),(\frac {-1}{\sqrt{10}},\frac 3{\sqrt{10}})\}.$ I'm going to represent $v$ in relation with the base $B$.

so then $$v=x'(\frac 3{\sqrt{10}},\frac 1{\sqrt{10}})+y'(\frac {-1}{\sqrt{10}},\frac 3{\sqrt{10}})$$

so we have a system with $2$ equations and $2$ unknowns:

$3x'-y'=x\sqrt{10}$

$x'+3y'=y\sqrt{10}$

and I get the solutions $x'=\frac{y+3x}{\sqrt{10}}$ and $y'=\frac{3y-x}{\sqrt{10}}.$ And when I introduce them into the equation, theoretically (based on the problem, also they gave me that specific base to work with) I should've a simpler curve equation, so then I get :

$$26y^2+96xy+8y-6x+54x^2=10.$$

which I have no idea how to represent. Am I doing something wrong?

Also: I know I can just isolate the $y$ and then represent it like that, but I have to use this method.

You have a cross term that can be eliminated as in this link: https://math.stackexchange.com/questions/1102328/rotating-a-conic-section-to-eliminate-the-xy-term — rlartiga, Oct 23 '18 at 19:39
$x'=\frac{y+3x}{\sqrt{10}},y'=\frac{x-3y}{\sqrt{10}}$ gives $-y^2-y+9x^2-1=0$. — Jan-Magnus Økland, Oct 23 '18 at 19:50
I think you dropped a sign. You should have a $+48 xy$ canceling with a $-48 xy$ — Doug M, Oct 23 '18 at 19:54
The given equation is for $x$ and $y.$ The wrong step was in uploading $x', y'$. You should upload $x=(3x'-y')/\sqrt{10}$ and $y=(x'+3y')/\sqrt{10}.$ — user376343, Oct 23 '18 at 20:01
@DougM Right... however when i obtain that, how do I really draw it? — C. Cristi, Oct 24 '18 at 06:54

Jan-Magnus Økland · Accepted Answer · 2018-10-24T11:37:54.087

Your basis is close to a basis that diagonalises$$M=\begin{pmatrix}8&3\\3&0\end{pmatrix}$$ $M$ has matrix of unit length eigenvectors $$P=\begin{pmatrix}\frac{3}{\sqrt{10}}&\frac1{\sqrt{10}}\\\frac1{\sqrt{10}}&-\frac{3}{\sqrt{10}}\end{pmatrix}$$ so that $$P^tMP=\begin{pmatrix}9&0\\0&-1\end{pmatrix},$$ transforming your equation $$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}8&3\\3&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}-\frac1{\sqrt{10}}&\frac{3}{\sqrt{10}}\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}-1=0$$ by $\begin{pmatrix}x\\y\end{pmatrix}=P\begin{pmatrix}x'\\y'\end{pmatrix}$ into $$\begin{pmatrix}x'&y'\end{pmatrix}\begin{pmatrix}9&0\\0&-1\end{pmatrix}\begin{pmatrix}x'\\y'\end{pmatrix}+\begin{pmatrix}0&-1\end{pmatrix}\begin{pmatrix}x'\\y'\end{pmatrix}-1=0$$ or $$9x'^2-y'^2-y'-1=0,$$ from which you can easily get by $x''=x', y''=y'+\frac12$ that $9x''^2-y''^2-\frac34=0$ or $$(\frac{x''}{\frac1{2\sqrt{3}}})^2-(\frac{y''}{\frac{\sqrt{3}}{2}})^2=1$$ which is in standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and is just a rotation and translation of your original hyperbola.

Now you can find the foci: $b^2=c^2-a^2\implies c^2=\frac56$. So $(x',y')=(\pm \sqrt{\frac{5}{6}},-\frac12)$ or $(x,y)= (\frac{\sqrt{30}-1}{2\sqrt{10}},\frac{10\sqrt{3}+9\sqrt{10}}{60})\approx(0.71,0.76)\text{ and }(-\frac{1+\sqrt{30}}{2\sqrt{10}},\frac{3}{2\sqrt{10}}-\frac1{2\sqrt{3}})\approx(-1.02,0.19)$ and use the definition to draw it using $2a=\frac1{\sqrt{3}}\approx 0.58$ and this video:

If you just want to plot it, use something like geogebra and the command

ImplicitCurve(8x^2+6x y-x/sqrt(10)+3y/sqrt(10)-1)

Draw the curve $8x^2+6xy-\frac x{\sqrt{10}}+3\frac y{\sqrt{10}}=1$

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