91Ó°ÊÓ

The equation is$y^2−x^2−4y+2x-6=0$

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

Simplify the equation.

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

Covert the equation in complete square

localid="1645783739225" $$\begin{gathered} \left(y^2−4y+4\right)-\left(x^2−2x+1\right)=6+4-1 \\ {\left(y-2\right)}^2-{\left(x-1\right)}^2=9 \end{gathered}$$

Divide the equation by 9 to get standard form of hyperbola

localid="1645784404156" $$\begin{gathered} \frac{{\left(y-2\right)}^2}{9}-\frac{{\left(x-1\right)}^2}{9}=\frac{9}{9} \\ \frac{{\left(y-2\right)}^2}{9}-\frac{{\left(x-1\right)}^2}{9}=1 \end{gathered}$$

In the equation $\frac{{\left(y-2\right)}^2}{9}-\frac{{\left(x-1\right)}^2}{9}=1$

The y² 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, $\left(h,k\right)=\left(1,2\right)$ and $a^2=9⇒a=Â\pm 3$and $b^2=9⇒b=Â\pm 3$

Sketch the rectangle that goes through the points 3 units above/below the centre 3 units left/right of centre

Sketch the asymptotes and draw the graph.