Asymptotes and focus of a conic? This is the conic $$x^2+6xy+y^2+2x+y+\frac{1}{2}=0$$
the matrices associated with the conic are:
$$
A'=\left(\begin{array}{cccc}
\frac{1}{2} & 1 & \frac{1}{2} \\
1 & 1 & 3 \\
\frac{1}{2} & 3 & 1
\end{array}\right),
$$
$$
A=\left(\begin{array}{cccc}
1 & 3 \\
3 & 1
\end{array}\right),
$$
His characteristic polynomial is: $p_A(\lambda) = \lambda^2-2\lambda-8$
A has eigenvalues discordant $(\lambda = 4, \lambda = -2)$, so it's an Hyperbole.
Then i found that the center of the conic is: $(-\frac{1}{16}, -\frac{5}{16})$
Then with the eigenvalues i found the two lines passing through the center:
$$4x-4y-1=0$$
$$8x+8y+3=0$$

Now i want to find the focus and the asymptotes but i have no idea how to do it.There is a way to find These two things through the data I have now? or do i need the canonical form of the conical? Thanks
 A: You've two principal axes:
\begin{align*}
  x^2+6xy+y^2+2x+y+\frac{1}{2} & \equiv
 A\left( \frac{4x-4y-1}{\sqrt{4^2+4^2}} \right)^2+
 B\left( \frac{8x+8y+3}{\sqrt{8^2+8^2}} \right)^2+C \\
 & \equiv
 -2\left( \frac{4x-4y-1}{4\sqrt{2}} \right)^2+
 4\left( \frac{8x+8y+3}{8\sqrt{2}} \right)^2+\frac{9}{32}
\end{align*}
Now
\begin{align*}
  \frac{4x-4y-1}{4\sqrt{2}} &= x'\\
  \frac{8x+8y+3}{8\sqrt{2}} &= y'\\
  a &= \frac{3}{8} \\
  b &= \frac{3}{8\sqrt{2}} \\
  \frac{x'^2}{a^2}-\frac{y'^2}{b^2} &=1
\end{align*}
Asymptotes
$$b x' \pm a y'=0$$
Foci
$$(x',y')=(\pm \sqrt{a^2+b^2},0)$$
A: For a hyperbola in standard position, $\tan\alpha=\frac b a$, where $\alpha$ is the angle that the asymptotes make with the transverse axis. If you transform the general equation into standard form, you’ll find that $a^2=-|A'|/\lambda_2|A|$ and $b^2=|A'|/\lambda_1|A|$, where $\lambda_1$ is the positive eigenvalue, so $\tan\alpha=\sqrt{-\lambda_2/\lambda_1}$. (Note that $|A|=\lambda_1\lambda_2$, so you don’t have to compute that separately.) With this value in hand, you can then use the formulas for tangents of the sum and difference of angle to find the slopes of the asymptotes.  
We have $\tan\alpha=\sqrt{2/4}=1/\sqrt2$, and the slope of the transverse axis is $-1$, so for the slopes of the asymptotes we get $${-1+1/\sqrt2\over1-(-1)\cdot1/\sqrt2}=-3+2\sqrt2$$ and $${-1-1/\sqrt2\over1+(-1)\cdot1/\sqrt2}=-3-2\sqrt2.$$ From this, we have $$y+\frac5{16}=(-3\pm2\sqrt2)\left(x+\frac1{16}\right)$$ for equations of the asymptotes, which you can then rearrange as you see fit.  
For the foci, we can use the fact that the eccentricity of a hyperbola is $e=\sqrt{1+b^2/a^2}=\sqrt{1-\lambda_2/\lambda_1}$ and that the distance from the center to the focus is $f=ea$. For this hyperbola, we have $$e=\sqrt{1+\frac24}=\frac{\sqrt6}2 \\ a^2=-{|A'|\over\lambda_2|A|}=-{-9/4\over(-2)(-8)}=\frac9{64}$$ so $$f=\frac{\sqrt6}2\cdot\frac38={3\sqrt6\over16}$$ which means that the foci are at $$\left(-\frac1{16},-\frac5{16}\right)\pm{3\sqrt6\over16}\left(\frac1{\sqrt2},-\frac1{\sqrt2}\right),$$ approximately $(0.262,-0.637)$ and $(-0.387,0.012)$.
A: These two examples are from a book that does not assume linear algebra...  Both good and bad, as they are giving concrete methods, which are a bit much to memorize. Worth doing both ways, really, get them to agree; this way, then linear algebra and coordinate changes (translation to center followed by rotation). They do asymptotes and foci below


