Question

CorrectFind the equation of the hyperbola with the following properties. Express your answer in standard form.Foci at (6,8) and (6,14)Vertices at (6,10) and (6, - 12)D

Answer 1

The standard form of the equation of the hyperbola is

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

(h, k) are the coordinates of the center

(h, k + a), (h, k - a) are the vertices

(h, k + c), (h, k - c) are the foci

Since the foci are (6, -8), (6, -14), then

$\begin{gathered} h=6 \\ k+c=-8 \\ k-c=-14 \end{gathered}$

Since the vertices are (6, -10), (6, -12), then

$\begin{gathered} h=c \\ k+a=-10 \\ k-a=-12 \end{gathered}$

To find k we will add the equations of k and a, or k and c

$\begin{gathered} k+a+k-a=-10+-12 \\ 2k=-22 \\ \frac{2k}{2}=\frac{-22}{2} \\ k=-11 \end{gathered}$

Substitute the value of k in the equation of foci to find and vertices to find a and c

$\begin{gathered} -11+c=-8 \\ -11+11+c=-8+11 \\ c=3 \end{gathered}$$\begin{gathered} -11+a=-10 \\ -11+11+a=-10+11 \\ a=1 \end{gathered}$

To find b we will use the equation

$c^2=a^2+b^2$

Since a = 1 and c = 3, then

$\begin{gathered} 3^2=1^2+b^2 \\ 9=1+b \\ 9-1=1-1+b^2 \\ 8=b^2 \end{gathered}$

Substitute the value of h, k, a^2, b^2 in the form of the equation above

$\begin{gathered} h=6 \\ k=-11 \\ a^2=1 \\ b^2=8 \\ \frac{(y--11)^2}{1}-\frac{(x-6)^2}{8}=1 \\ (y+11)^2-\frac{(x-6)^2}{8}=1 \end{gathered}$

The equation of the hyperbola in the standard form is

$(y+11)^2-\frac{(x-6)^2}{8}=1$