Chapter 8.6, Problem 44E

Solution

To classify the conic $x^2-4y^2+8x-24y-24=0$ . Then find the standard equation and graph it.

The conic is a hyperbola whose equation in standard form is $\frac{{\left(x+4\right)}^2}{2^2}-\frac{{\left(y+3\right)}^2}{1^2}=1$.

  

Given information:

The conic, $x^2-4y^2+8x-24y-24=0$.

Calculation:

Note that for the equation $x^2-4y^2+8x-24y-24=0$ , $A=1$ , $B=0$ and $C=-4$ . So, the value of the discriminant $B^2-4AC$ is:

  $\begin{array}{c}{B}^2-4AC=0^2-4\cdot 1\cdot \left(-4\right)\\ =16\end{array}$

Since $B^2-4AC>0$ , the conic is a hyperbola.

Now, rewrite the conic in standard form.

  $\begin{array}{c}{x}^2-4y^2+8x-24y-24=0\\ x^2+8x-4y^2-24y=24\\ x^2+8x+16-4\left(y^2-6y+9\right)+36=24+16\\ x^2+8x+16-4\left(y^2-6y+9\right)+36=24+16\\ {\left(x+4\right)}^2-4{\left(y+3\right)}^2=4\\ \frac{{\left(x+4\right)}^2}{2^2}-\frac{{\left(y+3\right)}^2}{1^2}=1\end{array}$

Therefore, the standard equation of the conic is $\frac{{\left(x+4\right)}^2}{2^2}-\frac{{\left(y+3\right)}^2}{1^2}=1$.

From the equation $\frac{{\left(x+4\right)}^2}{2^2}-\frac{{\left(y+3\right)}^2}{1^2}=1$ , the transverse axis is horizontal, $\left(h,k\right)=\left(-4,-3\right)$ , $a=2$ and $b=1$ . The vertices are at $\left(-4\pm a,-3\right)=\left(-4\pm 2,-3\right)$ or $\left(-6,-3\right)$ and $\left(-2,-3\right)$.

Plot the center and vertices. Then draw a rectangle $2a=2\times 2=4$ units wide and $2b=2\times 1=2$ units heigh centered at $\left(-4,-3\right)$ , draw the asymptotes and draw the hyperbola.