Chapter EP, Problem 10.6.12EP

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 10.6.12EP

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

Explanation of Solution

Given:

The equation is $4x^2-y^2=4$ .

Calculation:

Divide the above equation by 4.

$\begin{array}{c}\frac{4x^2}{4}-\frac{y^2}{4}=\frac{4}{4}\\ \frac{x^2}{1^2}-\frac{y^2}{2^2}=1\\ \frac{{\left(x-0\right)}^2}{1^2}-\frac{{\left(y-0\right)}^2}{2^2}=1\end{array}$

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 $4x^2-y^2=4$ or $\frac{{\left(x-0\right)}^2}{1^2}-\frac{{\left(y-0\right)}^2}{2^2}=1$ is as follows.

Figure (1)

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