Question

In the following exercises, a). Write the equation in standard form and b). graph.

$16x^2-4y^2+64x-24y-36=0$

Short answer

Part (a) The equation of hyperbola is$\frac{{\left(x+2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{16}=1$

Part (b) The graph is

Step-by-step solution

Part (a) Step 1. Given information

The equation is $16x^2-4y^2+64x-24y-36=0$

Write the equation in standard form of hyperbola and draw the graph.

Part (a) Step 2. Write the equation in standard form of hyperbola 

Simplify the equation.

$$\begin{gathered} 16x^2-4y^2+64x-24y-36=0 \\ 16\left(x^2+4x\right)-4\left(y^2+6y\right)=36 \end{gathered}$$

Covert the equation in complete square

$$\begin{gathered} 16\left(x^2+4x+2\right)-4\left(y^2+6y+9\right)=36+64-36 \\ 16{\left(x+2\right)}^2-4{\left(y+3\right)}^2=64 \end{gathered}$$

Divide the equation by 64 to get standard form of hyperbola

$$\begin{gathered} \frac{16{\left(x+2\right)}^2}{64}-\frac{4{\left(y+3\right)}^2}{64}=\frac{64}{64} \\ \frac{{\left(x+2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{16}=1 \end{gathered}$$

Part (b)  Step 1. Find centre and asymptote.

In the equation $\frac{{\left(x+2\right)}^2}{4}-\frac{{\left(y+3\right)}^2}{16}=1$,

The x² term is positive, the transverse axis is horizontal.

The hyperbola opens left and right .

Find the centre

Compare the equation with standard formlocalid="1645771717531" $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

So, $\left(h,k\right)=\left(-2,-3\right)$ and $a^2=4\Rightarrow a=\pm 2$ and localid="1645770960740" $b^2=16\Rightarrow b=\pm 4$

Sketch the rectangle that goes through the points 2 units left/right of centre 4 units above/ below the centre.

Part (b) Step 2. Draw the graph

Sketch the asymptotes and draw the graph