Chapter 8, Problem 27RE

Solution

To find: Identify the conic, write it in standard form by completing the square and sketch the graph.

The given conic is a hyperbola.

The standard form of the given conic is $\frac{{\left(y+4\right)}^2}{30}-\frac{{\left(x-3\right)}^2}{45}=1$ .

The graph of the given conic is shown in figure (1).

Given information:

The second degree equation is

  $2x^2-3y^2-12x-24y+60=0$ .

Formula used:

The second degree equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ graphs as

• Hyperbola if $B^2-4AC>0$ .
• Parabola if $B^2-4AC=0$ .
• Ellipse if $B^2-4AC<0$ .

Calculation:

Comparing the given equation $2x^2-3y^2-12x-24y+60=0$ with $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ gives

  $A=2$

  $B=0$

  $C=-3$

Then

  $\begin{array}{c}{B}^2-4AC=0-4\left(2\right)\left(-3\right)\\ B^2-4AC=24\\ B^2-4AC>0\end{array}$

Thus, the given conic is a hyperbola. Again considering the equation

  $2x^2-3y^2-12x-24y+60=0$

Isolate the constant term and completing the square

  $\begin{array}{c}2x^2-3y^2-12x-24y=-60\\ 2\left(x^2-6x\right)-3\left(y^2+8y\right)=-60\\ 2\left(x^2-6x+9\right)-3\left(y^2+8y+16\right)=-60+18-48\\ 2{\left(x-3\right)}^2-3{\left(y+4\right)}^2=-90\end{array}$

Simplifying the above equation

  $\begin{array}{c}2{\left(x-3\right)}^2-3{\left(y+4\right)}^2=-90\\ -\frac{{\left(x-3\right)}^2}{45}+\frac{{\left(y+4\right)}^2}{30}=1\\ \frac{{\left(y+4\right)}^2}{30}-\frac{{\left(x-3\right)}^2}{45}=1\end{array}$

This is the standard form of hyperbola with center $\left(h,k\right)$ .

Comparing it with $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ gives

  $\left(h,k\right)=\left(3,-4\right)$

Thus the center of the given hyperbola is $\left(3,-4\right)$

The graph of the equation $\frac{{\left(y+4\right)}^2}{30}-\frac{{\left(x-3\right)}^2}{45}=1$ is

  

Figure (1)