To find the standard form of the equation of a hyperbola given the vertices and foci, we must first determine several key parameters: the center [tex]\((h, k)\)[/tex], the distance [tex]\(a\)[/tex] (which is half the distance between the vertices), and the value [tex]\(c\)[/tex] (which is the distance from the center to each focus). We also need [tex]\(b^2\)[/tex], which we can find using the relationship [tex]\(c^2 = a^2 + b^2\)[/tex].
Given vertices [tex]\((5, 2)\)[/tex] and [tex]\((5, 10)\)[/tex]:
1. Calculate the Center [tex]\((h, k)\)[/tex]:
The center of the hyperbola is the midpoint of the segment connecting the vertices.
[tex]\[
h = \frac{x_1 + x_2}{2} = \frac{5 + 5}{2} = 5
\][/tex]
[tex]\[
k = \frac{y_1 + y_2}{2} = \frac{2 + 10}{2} = 6
\][/tex]
Therefore, the center is [tex]\((5, 6)\)[/tex].
2. Calculate [tex]\(a\)[/tex]:
The value [tex]\(a\)[/tex] is half the distance between the vertices:
[tex]\[
a = \frac{10 - 2}{2} = 4
\][/tex]
3. Calculate [tex]\(c\)[/tex]:
The value [tex]\(c\)[/tex] is the distance from the center to each focus. Given the foci [tex]\((5, 0)\)[/tex] and [tex]\((5, 12)\)[/tex]:
[tex]\[
c = |0 - 6| = 6 \quad \text{(or)} \quad c = |12 - 6| = 6
\][/tex]
4. Calculate [tex]\(b^2\)[/tex] using the relationship [tex]\(c^2 = a^2 + b^2\)[/tex]:
[tex]\[
c^2 = a^2 + b^2
\][/tex]
Substituting the known values,
[tex]\[
6^2 = 4^2 + b^2
\][/tex]
[tex]\[
36 = 16 + b^2
\][/tex]
[tex]\[
b^2 = 36 - 16
\][/tex]
[tex]\[
b^2 = 20
\][/tex]
5. Write the Standard Form of the Hyperbola:
Since the hyperbola has a vertical transverse axis (the vertices and foci have the same [tex]\(x\)[/tex]-coordinates), the standard form centered at [tex]\((h, k)\)[/tex] is:
[tex]\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
\][/tex]
Substituting [tex]\(h = 5\)[/tex], [tex]\(k = 6\)[/tex], [tex]\(a = 4\)[/tex], and [tex]\(b^2 = 20\)[/tex], we get:
[tex]\[
\frac{(y - 6)^2}{16} - \frac{(x - 5)^2}{20} = 1
\][/tex]
Therefore, the standard form of the equation of the hyperbola is:
[tex]\[
\boxed{\frac{(y - 6)^2}{16} - \frac{(x - 5)^2}{20} = 1}
\][/tex]
