Chapter 10, Problem 15RE

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

Hyperbola; center at $\left(-2,-3\right)$; focus at $$\left(-4,-3\right)$$; vertex at $$\left(-3,-3\right)$$

To determine

a.

To find: The equation of the conic described.

Answer to Problem 15RE

$\frac{{\left(x+2\right)}^2}{1}-\frac{{\left(y+3\right)}^2}{3}=1$

Explanation of Solution

Given:

Hyperbola; center at $\left(-2,-3\right)$; focus at $\left(-4,-3\right)$; vertex at $\left(-3,-3\right)$.

Formula used:

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

Calculation:

Hyperbola; center at $\left(-2,-3\right)$; focus at $\left(-4,-3\right)$; vertex at $\left(-3,-3\right)$.

Center $\left(h,k\right)=\left(-2,-3\right)$

$h=-2,k=-3$

Focus: $\left(h\pm c,k\right)=\left(-4,-3\right)$

$c=2$

Vertex: $\left(h\pm a,k\right)=\left(-3,-3\right)$

$a=1$

$b²=c²-a²$

$b²={\left(2\right)}^2-\left(1\right)²=3,b=\surd 3$

Equation: $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$,

$\frac{{\left(x+2\right)}^2}{1}-\frac{{\left(y+3\right)}^2}{3}=1$

Asymptote: $y-k=\pm \frac{b}{a}\left(x-h\right)$

$y+3=\pm \surd 3\left(x+2\right)$

To determine

b.

To graph: The equation $\frac{{\left(x+2\right)}^2}{1}-\frac{{\left(y+3\right)}^2}{3}=1$.

Answer to Problem 15RE

Explanation of Solution

Given:

Hyperbola; center at $\left(-2,-3\right)$; focus at $\left(-4,-3\right)$; vertex at $\left(-3,-3\right)$.

Graph:

The graph of the equation of $\frac{{\left(x+2\right)}^2}{1}-\frac{{\left(y+3\right)}^2}{3}=1$ is plotted.

Using the given information Hyperbola; center at $\left(-2,-3\right)$; focus at $\left(-4,-3\right)$; vertex at $\left(-3,-3\right)$.

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