Chapter 10, Problem 32RE

To determine

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

Answer to Problem 32RE

The hyperbolic equationis $\frac{{\left(x-1\right)}^{\text{2}}}{16}\text{-}\frac{{\left(y-3\right)}^{\text{2}}}{\text{20}}\text{=1}$ .

Explanation of Solution

Given:

The hyperbola vertices are $\left(\text{h,k}\right)$ = $\left(-3,3\right)\text{ and }\left(5,3\right)$ .

The focus is at $\left(7,3\right)$ .

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:

The hyperbola verticesare $\left(\text{h,k}\right)$ = $\left(-3,3\right)\text{ and }\left(5,3\right)$ .

The focus is at $\left(7,3\right)$ .

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

  $\left(\text{h±a,k}\right)\Rightarrow \text{h+a =5, h-a =-3 and k=3}\text{.}$ .

  $\text{Thus h=1, a=4, k=3}\text{.}$

Focus= $\left(\text{h±c,k}\right)=\left(7,3\right)$ .

Thus, $\text{c=6}\text{.}$

Now, $\text{b=}\sqrt{{\text{c}}^{\text{2}}{\text{-a}}^{\text{2}}}\text{=}\sqrt{\text{36-16}}\text{=}\sqrt{20}\text{.}$

Hence the equation is

  $\frac{{\left(x-1\right)}^{\text{2}}}{16}\text{-}\frac{{\left(y-3\right)}^{\text{2}}}{\text{20}}\text{=1}$ .

The equation can be in shown in graph as:

The graph of hyperbolic equation $\frac{{\left(x-1\right)}^{\text{2}}}{16}\text{-}\frac{{\left(y-3\right)}^{\text{2}}}{\text{20}}\text{=1}$ .

Hence, hyperbolic equation is $\frac{{\left(x-1\right)}^{\text{2}}}{16}\text{-}\frac{{\left(y-3\right)}^{\text{2}}}{\text{20}}\text{=1}$ .