Chapter EP, Problem 10.6.9EP

To determine

To write: The equation in standard form, state the graph of the equation is a parabola, circle, ellipse, or hyperbola and then graph the equation.

Answer to Problem 10.6.9EP

The standard form of equation is $\frac{{\left(x-\left(-5\right)\right)}^2}{6^2}-\frac{{\left(y-0\right)}^2}{3^2}=1$ , the equation is Hyperbola and graph is shown in figure (1).

Explanation of Solution

Given:

The equation is $x^2+10x+5=4y^2+16$ .

Calculation:

Rearrange the equation in the form of perfect square as follows.

$\begin{array}{c}{x}^2+10x+5=4y^2+16\\ x^2+2\left(5\right)x+25-25+5=4y^2+16\\ {\left(x+5\right)}^2-20=4y^2+16\\ {\left(x+5\right)}^2-4y^2=36\end{array}$

Divide the above equation by 36.

$\begin{array}{c}\frac{{\left(x+5\right)}^2}{36}-\frac{4y^2}{36}=\frac{36}{36}\\ \frac{{\left(x+5\right)}^2}{36}-\frac{y^2}{9}=1\\ \frac{{\left(x-\left(-5\right)\right)}^2}{6^2}-\frac{{\left(y-0\right)}^2}{3^2}=1\end{array}$

The general form of Hyperbola equation is $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

Therefore, the given equation is Hyperbola.

The graph of the equation $x^2+10x+5=4y^2+16$ or $\frac{{\left(x-\left(-5\right)\right)}^2}{6^2}-\frac{{\left(y-0\right)}^2}{3^2}=1$ is as follows.

Figure (1)

Therefore, the standard form of equation is $\frac{{\left(x-\left(-5\right)\right)}^2}{6^2}-\frac{{\left(y-0\right)}^2}{3^2}=1$ , the equation is Hyperbola and graph is shown in figure (1).