Chapter 8.6, Problem 40E

Solution

To find: The type of conic section, write equation in standard form and graph of the equation.

The equation $9y^2-x^2-54y+8x+56=0$ is a hyperbola and the standard form is $-\frac{{\left(x-4\right)}^2}{9}+\frac{{\left(y-3\right)}^2}{1}=1$ . The graph is shown in figure (1).

Given information: The given equation is $9y^2-x^2-54y+8x+56=0$.

Formula used: The general second-degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ describe any conic.

The type of conic is a circle if $B^2-4AC<0,B=0,A=C$.

The type of conic is an ellipse if $B^2-4AC<0,B\ne 0,A\ne C$.

The type of conic is a parabola if $B^2-4AC=0$.

The type of conic is a hyperbola if $B^2-4AC>0$.

Calculation:

Consider the given equation.

  $9y^2-x^2-54y+8x+56=0$

Compare the above equation with the general equation of the conic.

  $A=-1,B=0,C=9$

Substitute $-1$ for $A$ , $0$ for $B$ and $9$ for $C$ in $B^2-4AC$ to find the discriminant.

  $\begin{array}{c}{B}^2-4AC={\left(0\right)}^2-4\left(-1\right)\left(9\right)\\ =36\end{array}$

Since $B^2-4AC>0$ . So, the equation satisfies all the conditions of a hyperbola.

Rewrite the given equation in the standard form.

  $\begin{array}{c}-x^2+9y^2-54y+8x+56=0\\ -\left(x^2-8x+16\right)+16+9\left(y^2-6y+9\right)-81=-56\\ -{\left(x-4\right)}^2+9{\left(y-3\right)}^2=9\\ -\frac{{\left(x-4\right)}^2}{9}+\frac{{\left(y-3\right)}^2}{1}=1\end{array}$

The center of the hyperbola is $\left(4,3\right)$.

Drawn the graph of the ellipse using the above conditions.

  

Figure (1)

Therefore, the equation $9y^2-x^2-54y+8x+56=0$ is a hyperbola and the standard form is $-\frac{{\left(x-4\right)}^2}{9}+\frac{{\left(y-3\right)}^2}{1}=1$.