Chapter 10, Problem 12RE

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

Hyperbola; center at $\left(0,\text{ 0}\right)$; focus at $\left(0,\text{ 4}\right)$;vertex at $\left(0,-2\right)$

To determine

a.

To find: The equation of the conic described.

Answer to Problem 12RE

$\frac{y^2}{4}-\frac{x^2}{12}=1$

Explanation of Solution

Given:

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

Formula used:

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

Calculation:

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

Focus: $\left(0,\pm c\right)=\left(0,4\right)$

$c=4$

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

$a=2$

$b²=c²-a²$

$b²=4²-2²=12,b=2\surd 3$

Equation: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

$\frac{y^2}{4}-\frac{x^2}{12}=1$

Asymptote: $x=\pm \frac{b}{a}y=\pm \surd 3y$

To determine

b.

To graph: The equation $\frac{y^2}{4}-\frac{x^2}{12}=1$.

Answer to Problem 12RE

Explanation of Solution

Given:

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

Graph:

The graph of the equation of $\frac{y^2}{4}-\frac{x^2}{12}=1$ 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 $\frac{y^2}{4}-\frac{x^2}{12}=1$.