91Ó°ÊÓ

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.

localid="1645415725330" $$\begin{gathered} (4y^2-8y)+(-16x^2+32x)=76 \\ 4(y^2-2y)-16(x^2-2x)=76 \\ 4(y^2-2y+1)-16(x^2-2x+1)=76+4-16 \\ 4{(y-1)}^2-16{(x-1)}^2=64 \\ \frac{4{(y-1)}^2-16{(x-1)}^2}{64}=\frac{64}{64} \\ \frac{{(y-1)}^2}{16}-\frac{{(x-1)}^2}{4}=1 \end{gathered}$$

As $y^2$term is positive, so the transverse axis is vertical.

On comparing with standard form, we get,

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

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$(1,1)anda=4,b=2$

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

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

Thus, the graph of hyperbola is