Classify the conic $x^2+xy+3y^2+5x$ and determine its cartesian equation

Question

Classify the conic $C:\;x^2+xy+3y^2+5x$, and determine its cartesian equation

I would be happy if you solve it close to my way

A=\begin{pmatrix}1&\frac{1}{2}&\frac{5}{2}\\ \frac{1}{2}&3&0\\ \frac{5}{2}&0&0\end{pmatrix} Det(A)= -$\frac {75}{4}$ = $(α11)*(α22)*(α33)$ (≠0 irreducible)

A33= $\frac {75}{4}$ = $(α11)*(α22)$ (>0 elipse)

I = $α11+α22$

$α11X^2+α2Y^2+α33=0$

thats all I know

I could not find the cartesian equation

Answer 1

The rotation angle for $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is $\theta$ where $\tan{2\theta}=\frac{B}{A-C}$ which make $\cos{\theta}=\frac1{\sqrt{(2-\sqrt5)^2+1}},\sin{\theta}=-\frac1{\sqrt{(2+\sqrt5)^2+1}},$ for $x^2+xy+3y^2+5x=0.$ Let $M=\begin{pmatrix}1&\frac12\\\frac12&3\end{pmatrix},$ $P^tMP=\begin{pmatrix}(2-\frac{\sqrt5}2)&0\\0&(2+\frac{\sqrt5}2)\end{pmatrix},$ where $P=\begin{pmatrix}\cos{\theta}&-\sin{\theta}\\\sin{\theta}&\cos{\theta} \end{pmatrix}.$ The original equation therefore can be written $$\frac{75}{11}(\frac{(\cos{(\theta)}(x+\frac{30}{11})+\sin{(\theta)}(y-\frac5{11}))^2}{a^2}+\frac{(-\sin{(\theta)}(x+\frac{30}{11})+\cos{(\theta)}(y-\frac5{11}))^2}{b^2}-1)=0.$$ where $a^2=\frac{75}{11(2-\frac{\sqrt5}2)}\approx (2.78)^2$ and $b^2=\frac{75}{11(2+\frac{\sqrt5}2)}\approx (1.48)^2,$ from which you should be able to extract what you need. Even the area enclosed $\pi ab=\frac{150\pi}{11\sqrt{11}}.$ Or the squared eccentricity $(1-\frac{b^2}{a^2})=\frac2{11}(4\sqrt5-5),$ making the eccentricity $\approx 0.85.$ The foci, directrices and vertices may require some more work.

Answer 2

The conic matrix $$A=\begin{pmatrix}1 & \frac{1}{2} & \frac{5}{2}\\ \frac{1}{2} & 3 & 0\\ \frac{5}{2} & 0 & 0 \end{pmatrix} \tag{1}$$

can be transformed into the following diagonal form

$$(T\;R)^{\intercal}\;A\;(T\;R)=\begin{pmatrix}2-\frac{\sqrt{5}}{2}\\ & 2+\frac{\sqrt{5}}{2}\\ & & \text{-}\frac{75}{11} \end{pmatrix} \tag{2}$$

using the translation matrix

$$T=\begin{pmatrix}1 & & \text{-}\frac{30}{11}\\ & 1 & \frac{5}{11}\\ & & 1 \end{pmatrix} \tag{3}$$

and rotation matrix

$$R=\begin{pmatrix}\cos\left(\tfrac{1}{2}{\rm atan}\left(\tfrac{1}{2}\right)\right) & \text{-}\sin\left(\tfrac{1}{2}{\rm atan}\left(\tfrac{1}{2}\right)\right)\\ \sin\left(\tfrac{1}{2}{\rm atan}\left(\tfrac{1}{2}\right)\right) & \cos\left(\tfrac{1}{2}{\rm atan}\left(\tfrac{1}{2}\right)\right)\\ & & 1 \end{pmatrix} \tag{4}$$

From (2) the canonical form $\left(\tfrac{u}{a}\right)^{2}+\left(\tfrac{v}{b}\right)^{2}=1$ is

$$\begin{aligned}a=\sqrt{\tfrac{600}{121}+\tfrac{150\sqrt{5}}{121}} & & \text{semi-major axis}\\ b=\sqrt{\tfrac{600}{121}-\tfrac{150\sqrt{5}}{121}} & & \text{semi-minor axis} \end{aligned} \tag{5}$$

At this point you can evaluate the eccentricity $\epsilon = \sqrt{1 - \left( \tfrac{b}{a} \right)^2}$ and the focii distance $f = \epsilon a$.