Chapter 9.3, Problem 50E

To determine

To find: The standard form of the equation of hyperbola with the given characteristics.

Answer to Problem 50E

The standard form of the equation of hyperbola is $\frac{{\left(y-1\right)}^2}{4}-\frac{{\left(x-4\right)}^2}{1}=1$ .

Explanation of Solution

Given information:

The characteristics are foci: $\left(1,2\right),\left(1,6\right)$ ; asymptotes: $y=2+2x,y=6-2x$ .

Calculation:

For a up-down facing hyperbola, the Foci are defined as $\left(k,h+c\right)$ and $\left(k,h-c\right)$ .

Solve for center $\left(h,k\right)$ by comparing with the coordinates of given foci.

  $h+c=2$… (1)

  $h-c=6$.........(2)

  $k=1$

Add equation (1) and (2) to find $h$ .

  $\begin{array}{c}\left(h+c\right)+\left(h-c\right)=2+6\\ 2h=8\\ h=\frac{8}{2}\\ =4\end{array}$

The center of the hyperbola will be $\left(4,1\right)$ .

Solve for $a$ and b by comparing the equation to the two given asymptotes.

  $\begin{array}{c}a=2\\ b=1\end{array}$

Substitute 4 for $h$ , 1 for $k$ , 2 for $a$ and 1 for $b$ in the equation $\frac{{\left(y-k\right)}^2}{a^2}-\frac{{\left(x-h\right)}^2}{b^2}=1$ to find the standard form of hyperbola equation.

  $\begin{array}{c}\frac{{\left(y-1\right)}^2}{2^2}-\frac{{\left(x-4\right)}^2}{1^2}=1\\ \frac{{\left(y-1\right)}^2}{4}-\frac{{\left(x-4\right)}^2}{1}=1\end{array}$

Therefore, the standard form of the equation of the hyperbola is $\frac{{\left(y-1\right)}^2}{4}-\frac{{\left(x-4\right)}^2}{1}=1$ .