91影视

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="1645383547234" $$\begin{gathered} (16x^2-64x)-(4y^2+24y)=36 \\ 16(x^2-4x)-4(y^2+6y)=36 \\ 16(x^2-4x+2^2)-4(y^2+6y+3^2)=36-36+64 \\ 16{(x-2)}^2-4{(y+3)}^2=64 \\ \frac{16{(x-2)}^2-4{(y+3)}^2}{64}=\frac{64}{64} \\ \frac{{(x-2)}^2}{4}-\frac{{(y+3)}^2}{16}=1 \end{gathered}$$

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

On comparing with standard form, we get,

$(h,k)=(2,-3)$

And

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

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

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

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

The graph of hyperbola is