Chapter 10, Problem 36RE

To determine

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

Answer to Problem 36RE

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

Explanation of Solution

Given:

Hyperbola vertices are $\left(4,0\right)\text{ and }\left(4,4\right)$ .

Asymptote the line $\text{y+2x=10}$ .

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 $\left(4,0\right)\text{ and }\left(4,4\right)$ .

Asymptote the line $\text{y+2x=10}$ .

Thus, 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 $\left(\text{h-a,k}\right)=\left(4,0\right)$ and $\left(\text{h+a,k}\right)=\left(4,4\right)$ .

Asymptote the line $\text{y+2x=10}$ .

  $\text{y+2x=10}$ .

  $\text{y=10-2x}$ .

Thus, $\frac{\text{b}}{\text{a}}$ = $\frac{-2}{1}$ .

  $\text{h=3, k=4,a=1, b=-2}$ .

Hence the equation is

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

The equation can be in shown in graph as:

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

  

In the graph:

At $\text{y=0}$ ; $\text{x=0}\text{.764 and x=5}\text{.236}$ .

At $\text{x=0; y=9}\text{.657 and y=-1}\text{.657}$ .

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