<s>[INST] Conic Sections Question - Hyperbolas & Circles 
*

*So, if you have a hyperbola with foci at $(4,0)$ & $(-2,0)$, and the slopes of the asymptotes are $+4$ and $-4$, what would the equation for this hyperbola be? I know that the center would be $(1,0)$, and that would mean that $c$ is $3$ (and $c^2$ is $9$), but I don't know how to get a or be now. 

*You're given $2x^2 + 4x + 2y^2 + 6y = 66$, how would you find the center and the radius? I was thinking maybe complete the square, but for some reason it doesn't seem to be working correctly? 
If you could just give me some help on these two, that would be very much appreciated! Thank you so much. 
 [/INST] 1.  You are correct that the center is at $(1, 0)$, and so the focal length is indeed $c = 3$.
Recall that you can draw a rectangle, centered at the center of the hyperbola such that the asymptotes pass through the corners and the left and right sides of the rectangle are tangent to the vertices of the hyperbola.  The half-dimensions of the box are $a$ and $b$, and we know that
$$
c^2 = a^2 + b^2.
$$
Now, call the slope of the (positively sloped) asymptote $m$.  In this problem, $m = 4$.  Then,
$$
\frac{b}{a} = m \qquad \text{so} \qquad b = ma.
$$
Substitute back into the pythagorean equation to obtain:
$$
\begin{align}
c^2 &= a^2 + (ma)^2 \\
&= a^2 + m^2 a^2 \\
&= (1 + m^2) a^2
\end{align}
$$
And so,
$$
a^2 = \frac{c^2}{1 + m^2} \qquad \Longrightarrow \qquad a = \frac{c}{\sqrt{1 + m^2}}
$$
and
$$
b = \frac{mc}{\sqrt{1 + m^2}}
$$
You can find the values of $a$ and $b$ for your particular hyperbola.

2.  It will be easier to complete the square in $x$ and $y$ if you first make the coefficients of $x^2$ and $y^2$ equal to $1$.  This can be accomplished by first dividing by $2$.
$$
\begin{align}
2x^2 + 4x + 2y^2 + 6y &= 66 \\
x^2 + 2x + y^2 + 3y &= 33 \\
x^2 + 2x + 1 + y^2 + 3y + \frac{9}{4} &= 33 + 1 + \frac{9}{4} \\
(x + 1)^2 + \left( y + \frac{3}{2} \right)^2 = \frac{145}{4}
\end{align}
$$
From there, you can read off the center and radius.
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