Chapter 10, Problem 18RE

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

Vertices at $\left(0,1\right)$ and $\left(6,1\right)$; asymptote the line $3y+2x=9$

To determine

a.

To find: The equation of the conic described.

Answer to Problem 18RE

$\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{4}=1$

Explanation of Solution

Given:

Vertices at $\left(0,1\right)$ and $\left(6,1\right)$, asymptote the line $3y+2x=9$.

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:

Vertices at $\left(0,1\right)$ and $\left(6,1\right)$, asymptote the line $3y+2x=9$.

The curve is a hyperbola.

Vertices $\left(h\pm a,k\right)$ $=\left(0,1\right)$ and $\left(6,1\right)$.

$h=3, a=3, k=1$

Center $\left(h,k\right)=\left(3,1\right)$.

asymptote the line $3y+2x=9$.

The slope of thee asymptote is $-2/3$.

$-\frac{b}{a}=-2/3$;$b=2$;$a=3$

$b^2=c^2-a^2$;$c²=4+9=\surd 13$

Focus: $\left(h\pm c,k\right)=\left(3+\surd 13,1), (3-\surd 13, 1\right)$.

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

$\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{4}=1$

Asymptote: $\left(y-1\right)=\pm 2/3\left(x-3\right)$.

$3y-2x=-3$

To determine

b.

To graph: The equation $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{4}=1$.

Answer to Problem 18RE

Explanation of Solution

Given:

Vertices at $\left(0,1\right)$ and $\left(6,1\right)$, asymptote the line $3y+2x=9$.

Graph:

The equation $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{4}=1$.

Using the given information Vertices at $\left(0,1\right)$ and $\left(6,1\right)$, asymptote the line $3y+2x=9$.

We find, the center and $a=3$, $b=2$. We have formed the equation and plotted the graph for $\frac{{\left(x-3\right)}^2}{9}-\frac{{\left(y-1\right)}^2}{4}=1$.