Chapter 10.4, Problem 31AYU

In Problems 29-36, find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation by hand. Verify your graph using a graphing utility.

$4x^2-y^2=16$

To determine

a.

To find: Center, transverse axis, vertices, foci and asymptotes.

Answer to Problem 31AYU

Solution:

Center: $\left(0,0\right)$.

The transverse axis is $x\text{-axis}$.

Vertex: $(2,0)$, $(-2,0)$.

Foci: $(2\surd 5,0),(-2\surd 5,0)$.

Asymptotes: $y=2x,y=-2x$.

Explanation of Solution

Given:

$4x^2-y^2=16$

Formula used:

Center	Transverse axis	Foci	Vertices	Equation	Asymptotes
$(0,0)$	$x\text{-axis}$	$\left(\pm c,0\right)$	$\left(\pm a,0\right)$	$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1;b^2 = c^2 - a^2$	$y=\frac{b}{a}x;y=-\frac{b}{a}x$

Calculation:

Consider the given equation $4x^2-y^2=16$.

Rewriting the equation in the hyperbola form, we get,

$\frac{x^2}{4}-\frac{y^2}{16}=1$

The center of the hyperbola is $(0,0)$.

From the given data we see that, transverse axis is $x\text{-axis}$.

$a=2,b=4$ $c=\sqrt{b^2+a^2},c=2\surd 5$

Vertex: $\left(\pm a,0\right)=\left(\pm 2,0\right)$.

Foci: $\left(\pm c,0\right)=\left(\pm 2\surd 5,0\right)$.

Asymptotes: $y=\pm \frac{b}{a}x=\pm \frac{4}{2}x=\pm 2x$.

To determine

b.

To graph: The equation $\frac{x^2}{4}-\frac{y^2}{16}=1$.

Answer to Problem 31AYU

Solution:

Explanation of Solution

Given:

$4x^2-y^2=16$

Graph:

The graph of the equation of $\frac{x^2}{4}-\frac{y^2}{16}=1$ is plotted.

Using the given equation $4x^2-y^2=16$. We find $a=2$, $b=4$. We have found the center: $(0,0)$.

The transverse axis is $x\text{-axis}$.

Vertex: $\left(\pm 2,0\right)$.

Foci: $\left(\pm 2\surd 5,0\right)$.

Asymptotes: $y=\pm 2x$.