In the following exercises,

(a) write the equation in standard form and

(b) graph.

$4x^2-16y^2+8x+96y-204=0$

The given equation is$4x^2-16y^2+8x+96y-204=0$

From the given equation, it is clear that $x^{2}and{y}^2$term have different sign and different coefficient. So, the equation represents a hyperbola.

Now, we will group the x and y term and make the coefficient of $x^{2}and{y}^2$equal to 1 and then complete the squares and write as binomial.

role="math" localid="1645382378675" $$\begin{gathered} (4x^2+8x)+(-16y^2+96y)=204 \\ 4(x^2+2x)-16(y^2-6y)=204 \\ 4(x^2+2x+1)-16(y^2-6y+9)=204+4-144 \\ 4{(x+1)}^2-16{(y-3)}^2=64 \\ \frac{4{(x+1)}^2-16{(y-3)}^2}{64}=\frac{64}{64} \\ \frac{{(x+1)}^2}{16}-\frac{{(y-3)}^2}{4}=1 \end{gathered}$$

As $x^2$term is positive, so the transverse axis is horizontal.

On comparing with standard form, we get,

$(h,k)=(-1,3)$

And

$$\begin{gathered} a^2=16 \\ a=\sqrt{16} \\ a=4 \\ And \\ {b}^2=4 \\ b=\sqrt{4} \\ b=2 \end{gathered}$$

Thus, center is $(-1,3)anda=4,b=2$

Next, we will find the equation of asymptote as follows,

$$\begin{gathered} y=\frac{b}{a}(x-h)+k \\ y=\frac{2}{4}(x+1)+3 \\ y=\frac{1}{2}x+\frac{1}{2}+3 \\ y=\frac{1}{2}x+\frac{7}{2} \\ And \\ y=-\frac{b}{a}(x-h)+k \\ y=-\frac{2}{4}(x+1)+3 \\ y=-\frac{1}{2}x-\frac{1}{2}+3 \\ y=-\frac{1}{2}x+\frac{5}{2} \end{gathered}$$

The graph of hyperbola is