Chapter 10, Problem 17RE

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

Center at $\left(-1,2\right)$; $a\text{ = 3}$; $c\text{ = 4}$; transverse axis parallel to the $x\text{-axis}$.

To determine

a.

To find: The equation of the conic described.

Answer to Problem 17RE

$\frac{{\left(x+1\right)}^2}{9}-\frac{{\left(y-2\right)}^2}{7}=1$

Explanation of Solution

Given:

Center at $\left(-1,2\right)$; $a=3$; $c=4$; transverse axis parallel to the $x\text{-axis}$.

Formula Used:

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

Calculation:

Center at $\left(-1,2\right)$; $a=3$; $c=4$; transverse axis parallel to the $x\text{-axis}$.

The curve is a hyperbola.

Center $\left(h,k\right)$$=\left(-1,2\right)$.

$h=-1, k=2$

$c=4$

Focus: $\left(h\pm c,k\right)\text{ = }\left(3,2\right),\left(-5,\text{ 2}\right)$.

$a=3$

Vertex: $\left(h\pm a,k\right)\text{ = }\left(-4,2\right),\left(2,\text{ 2}\right)$.

$b²=c²-a²$

$b²={\left(4\right)}^2-{\left(3\right)}^2=7, b=\surd 7$

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

$\frac{{\left(x+1\right)}^2}{9}-\frac{{\left(y-2\right)}^2}{7}=1$

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

$y-2=\pm \sqrt{7}/3\left(x+1\right)$