Hyperbola: Standard Equation

Step 1: Determine the following:

Orientation of major axis: Since the two foci and two vertices fall on the horizontal line y = 8, the transverse axis is horizontal.

Center: The transverse axis is the line segment that connects the two vertices. Since the center is the midpoint of the transverse axis then it is also the midpoint of the two vertices.

$\left(h, k\right)=\left(\frac{-1+5}{2}, \frac{8+8}{2}\right)=\left(\frac{4}{2},\frac{16}{2}\right)=\left(2, 8\right)$

Length of a: Take one of the vertices and determine the length to the center.

Foci (-1, 8): $a=\left|2-\left(-1\right)\right|=\left|3\right|=3$

Foci (5, 8): $a=\left|5- 8\right|=\left|-3\right|=3$

a = 3

Length of b: To find b the equation $b=\sqrt{c^2-a^2}$ can be used but the value of c must be determined. Since c is the distance from the foci to the center, take either foci and determine the distance to the center. Then solve for b.

Foci (-3, 8): $c=\left|2-\left(-3\right)\right|=\left|5\right|=5$

Foci (7, 8): $c=\left|2- 7\right|=\left|-5\right|=5$

c = 5

$b=\sqrt{c^2-a^2}$

$b=\sqrt{5^2-3^2}=\sqrt{16}=4$

b = 4

Step 2: Substitute the values for h, k, a and b into the equation for a hyperbola with a transverse axis.

Equation for a horizontal transverse axis:

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

Substitute:

$\frac{{\left(x-2\right)}^2}{3^2}-\frac{{\left(y-8\right)}^2}{4^2}=1$

$\frac{{\left(x-2\right)}^2}{3^2}-\frac{{\left(y-8\right)}^2}{4^2}=1$

Step 1: Determine the following:

Orientation of major axis: Since the two foci fall on the vertical line x = 3, the transverse axis is vertical.

Center: The center is the midpoint of the two foci.

$\left(h, k\right)=\left(\frac{3+3}{2}, \frac{-5+5}{2}\right)=\left(\frac{6}{2},\frac{0}{2}\right)=\left(3, 0\right)$

Length of a: The value of a is the distance between the center and the vertices. Since given the distance from the vertices to the foci of one the value of a can be determined by finding the distance from the foci to the center and subtracting one. To find c take either foci and calculate the distance to the center. Then solve for a.

Foci (3, -5): $c=\left|0-\left(-5\right)\right|=\left|5\right|=5$

Foci (3, 5): $c=\left|0- 5\right|=\left|-5\right|=5$

c = 5

$a=c-distance from vertex to foci$

$a=5-1\to a=4$

Length of b: To find b the equation $b=\sqrt{c^2-a^2}$ can be used.

$b=\sqrt{c^2-a^2}$

$b=\sqrt{5^2-4^2}=\sqrt{9}=3$

b = 3

Step 2: Substitute the values for h, k, a and b into the equation for a hyperbola with a vertical transverse axis.

Equation for a vertical transverse axis:

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

Substitute:

$\frac{{\left(y-0\right)}^2}{4^2}-\frac{{\left(x-3\right)}^2}{3^2}=1$

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