Chapter 10, Problem 35RE

To determine

Tofind:the equation of conic using its given description and graph the equation.

Answer to Problem 35RE

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

Explanation of Solution

Given:

Hyperbola vertices are $\left(0,1\right)\text{ and }\left(6,1\right)$ .

Asymptote the line $\text{3y+2x=9}$ .

Concept used:

Formula for hyperbola forming open left and right.

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

Center= $\left(\text{h,k}\right)$ .

Vertices= $\left(\text{h+a,k}\right)$ and $\left(\text{h-a,k}\right)$ .

Slope of asymptotes: $\pm \frac{\text{b}}{a}$ .

Equation of asymptotes: $\text{y=k±}\frac{\text{b}}{\text{a}}\left(\text{x-h}\right)$ .

Formula for hyperbola opening up and down:

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

Center: $\left(\text{h,k}\right)$ .

Vertices: $\left(\text{h,k+b}\right)\text{ and }\left(\text{h,k-b}\right)$ .

Slope of asymptotes: $\text{±}\frac{\text{b}}{\text{a}}$ .

Calculation:

According to the given:

vertices are $\left(0,1\right)\text{ and }\left(6,1\right)$ .

0Thus, the equation is in the form:

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

vertices $\left(\text{h-a,k}\right)=\left(0,1\right)$ and $\left(\text{h+a,k}\right)=\left(6,1\right)$ .

Asymptote the line $\text{3y+2x=9}$ .

  $\text{3y=-2x+9}$ .

  $\text{y=-}\frac{\text{2}}{\text{3}}\text{x+3}$ .

Thus, $\frac{\text{b}}{\text{a}}$ = $\frac{2}{3}$ .

  $\text{h=3, k=1,a=3, b=2}$

Hence the equation is

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

The equation can be in shown in graph as:

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

  

In the graph:

At $\text{y=0}$ ; $\text{x=-0}\text{.354 and x=6}\text{.354}$ .

At $\text{x=0; y=1}$ .

Hence the equation $\frac{{\left(\text{x-3}\right)}^{\text{2}}}{9}\text{-}\frac{{\left(\text{y-1}\right)}^{\text{2}}}{4}\text{=1}$ is hyperbolic.