91Ó°ÊÓ

The equation is $4y^2-16x^2-24y+96x-172=0$.

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

Simplify the equation.

$$\begin{gathered} 4y^2-16x^2-24y+96x-172=0 \\ 4\left(y^2-6y\right)-16\left(x^2-6x\right)=172 \end{gathered}$$

Covert the equation in complete square

role="math" localid="1645980061372" $$\begin{gathered} 4\left(y^2-6y+9\right)-16\left(x^2-6x+9\right)=172+36-144 \\ 4{\left(y-3\right)}^2-16{\left(x-3\right)}^2=64 \end{gathered}$$

Divide the equation by 64 to get standard form of hyperbola

role="math" localid="1645980133774" $$\begin{gathered} \frac{4{\left(y-3\right)}^2}{64}-\frac{16{\left(x-3\right)}^2}{64}=\frac{64}{64} \\ \frac{{\left(y-3\right)}^2}{16}-\frac{{\left(x-3\right)}^2}{4}=1 \end{gathered}$$

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

They² term is positive, the transverse axis is vertical.

The hyperbola opens up and down.

Find the centre

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

So, $\left(h,k\right)=\left(3,3\right)$and role="math" localid="1645980920697" $a^2=16⇒a=Â\pm 4$and $b^2=4⇒b=Â\pm 2$

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

Sketch the asymptotes and draw the graph.