Chapter 10, Problem 13RE

In Problems 11-18, find an equation of the conic described. Graph the equation by hand.

Ellipse; foci at $\left(-3,\text{ 0}\right)$ and $\left(3,\text{ 0}\right)$; vertex at $\left(4,\text{ 0}\right)$

To determine

a.

To find: The equation of the conic described.

Answer to Problem 13RE

$\frac{x^2}{16}+\frac{y^2}{7}=1$

Explanation of Solution

Given:

Ellipse; foci at $\left(-3,0\right)$ and $\left(3,0\right)$; vertex at $\left(4,0\right)$.

Formula used:

Center	Major axis	Foci	Vertices	Equation
$\left(0,0\right)$	$x\text{-axis}$	$(\pm c,0)$	$(\pm a,0)$	$\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$ $b²=a²-c²$

Calculation:

Ellipse; foci at $\left(-3,0\right)$ and $\left(3,0\right)$; vertex at $\left(4,0\right)$.

Focus: $\left(\pm c,0\right)=\left(\pm 3,0\right)$

$c=3$

Vertex: $\left(a,0\right)=\left(4,0\right)$

$a=4$

$b²=a²-c²$

$b²=4²-3²=7,b=\surd 7$

Equation:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{16}+\frac{y^2}{7}=1$

To determine

b.

To graph: The equation $\frac{x^2}{16}+\frac{y^2}{7}=1$.

Answer to Problem 13RE

Explanation of Solution

Given:

Hyperbola; center at foci at $\left(-3,0\right)$ and $\left(3,0\right)$; vertex at $\left(4,0\right)$.

Graph:

The graph of the equation of $\frac{x^2}{16}+\frac{y^2}{7}=1$ is plotted.

Using the given information Hyperbola; center at $\left(0,0\right)$; focus at $\left(0,4\right)$; vertex at $\left(0,\mathrm{-2}\right)$.

We find $a=2,b=2\surd 3$. We have formed the equation and plotted the graph for $\frac{x^2}{16}+\frac{y^2}{7}=1$.