To find the standard form of the equation of the hyperbola with the given vertices and a point it passes through, we need to follow these steps:
1. Determine the center of the hyperbola:
The vertices of the hyperbola are [tex]\((1, -2)\) and \((1, -6)\).[/tex]
The center [tex]\((h, k)\)[/tex] is the midpoint of the vertices. So,
[tex]\[ h = \frac{1 + 1}{2} = 1 \] \[ k = \frac{-2 + (-6)}{2} = \frac{-8}{2} = -4 \][/tex]
Thus, the center of the hyperbola is [tex]\((1, -4)\).[/tex]
2. Determine the distance between the vertices:
The distance between the vertices is [tex]\(2a\)[/tex] (where [tex]\(a\)[/tex] is the distance from the center to each vertex).
[tex]\[ 2a = |-2 - (-6)| = |-2 + 6| = |4| = 4 \implies a = 2 \][/tex]
3. Determine the orientation of the hyperbola:
Since the vertices have the same x-coordinate and different y-coordinates, the hyperbola opens vertically.
The standard form of the equation for a vertically oriented hyperbola is:
[tex]\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \][/tex]
Substituting [tex]\(h = 1\), \(k = -4\), and \(a = 2\),[/tex] we get:
[tex]\[ \frac{(y + 4)^2}{4} - \frac{(x - 1)^2}{b^2} = 1 \][/tex]
4. Determine [tex]\(b^2\)[/tex] using the given point:
The hyperbola passes through the point [tex]\((5, -10)\).[/tex]
Substitute [tex]\((x, y) = (5, -10)\)[/tex] into the equation:
[tex]\[ \frac{(-10 + 4)^2}{4} - \frac{(5 - 1)^2}{b^2} = 1 \] \[ \frac{(-6)^2}{4} - \frac{(4)^2}{b^2} = 1 \] \[ \frac{36}{4} - \frac{16}{b^2} = 1 \] \[ 9 - \frac{16}{b^2} = 1 \] \[ 8 = \frac{16}{b^2} \] \[ b^2 = \frac{16}{8} = 2 \][/tex]
5. Write the standard form of the equation:
Substitute [tex]\(a^2 = 4\) and \(b^2 = 2\)[/tex] into the standard form equation:
[tex]\[ \frac{(y + 4)^2}{4} - \frac{(x - 1)^2}{2} = 1 \][/tex]
Thus, the standard form of the equation of the hyperbola is:
[tex]\[\boxed{\frac{(y + 4)^2}{4} - \frac{(x - 1)^2}{2} = 1}\][/tex]