Chapter 10.6, Problem 6CFU

To determine

To identify: the conic section represented by each equation; write the equation in standard form and then graph the equation.

Answer to Problem 6CFU

Hyperbola

  $\frac{{\left(x-2\right)}^2}{{\left(\sqrt{5}\right)}^2}-\frac{{\left(y-\left(-2\right)\right)}^2}{{\left(\sqrt{5}\right)}^2}=1$

Explanation of Solution

Given information:

  $x^2-4x-y^2-5-4y=0$

Calculation:

Rewrite the equation in the general form $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ to derive the coefficients as shown below.

  $\begin{array}{l}1x^2+0xy+\left(-1\right)y^2+\left(-4\right)x+\left(-4\right)y+\left(-5\right)=0\\ \therefore A=1\text{ and }C=-1\end{array}$

Since $A=-C$ , thus the conic section represented by the equation is a hyperbola.

The standard equation of a hyperbola is shown below.

  $\begin{array}{c}\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1\\ or\\ \frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1\\ or\\ xy=k\end{array}$

Now rewrite the equation in the standard form as follows:

  $\begin{array}{c}{x}^2-4x-y^2-5-4y=0\\ x^2-4x-y^2-4y=5\left[\text{Rewrite the equation}\right]\\ \left(x^2-4x+4\right)-\left(y^2+4y-4\right)=5+4-4\\ {\left(x-2\right)}^2-{\left(y-\left(-2\right)\right)}^2={\left(\sqrt{5}\right)}^2\\ \frac{{\left(x-2\right)}^2}{{\left(\sqrt{5}\right)}^2}-\frac{{\left(y-\left(-2\right)\right)}^2}{{\left(\sqrt{5}\right)}^2}=1\left[\text{Divide each side by 5}\right]\end{array}$

Thus, the standard equation of hyperbola is $\frac{{\left(x-2\right)}^2}{{\left(\sqrt{5}\right)}^2}-\frac{{\left(y-\left(-2\right)\right)}^2}{{\left(\sqrt{5}\right)}^2}=1$ .

The value of $h,k,a\text{ and }b$ is 2, $-2,\sqrt{5}$ and $\sqrt{5}$ respectively. The vertex $\left(h,k\right)$ of hyperbola is $\left(2,-2\right)$ and semi axis $\left(a\right)$ is $\sqrt{5}$ and semi conjugate axis $\left(b\right)$ is $\sqrt{5}$ . Thus, the graph of the given equation is shown below.