Answer :
To find the equation of the hyperbola based on the given vertices and foci, follow these steps:
1. Determine the center of the hyperbola:
The vertices are located at \((3, -6)\) and \((3, 6)\). The center of the hyperbola is the midpoint of the vertices.
Midpoint [tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{3 + 3}{2}, \frac{-6 + 6}{2} \right) = (3, 0) \][/tex]
2. Determine the distance between the vertices:
The distance between the two vertices is the length of the transverse axis, which is \(2a\). From \((3, -6)\) to \((3, 6)\), the distance is 12 units.
[tex]\[ 2a = 12 \implies a = \frac{12}{2} = 6 \][/tex]
3. Determine the distance between the foci:
The foci are located at \((3, -10)\) and \((3, 10)\). The distance between the foci is \(2c\). From \((3, -10)\) to \((3, 10)\), the distance is 20 units.
[tex]\[ 2c = 20 \implies c = \frac{20}{2} = 10 \][/tex]
4. Determine \(b\):
Use the relationship between \(a\), \(b\), and \(c\) for hyperbolas:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Substituting the known values:
[tex]\[ 10^2 = 6^2 + b^2 \implies 100 = 36 + b^2 \implies b^2 = 64 \][/tex]
5. Write the equation of the hyperbola:
Since the hyperbola opens vertically (as indicated by the vertical difference in the vertices and foci), its standard form is:
[tex]\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \][/tex]
Here, \(h = 3\), \(k = 0\), \(a = 6\), and \(b^2 = 64\):
[tex]\[ \frac{(y - 0)^2}{6^2} - \frac{(x - 3)^2}{64} = 1 \implies \frac{y^2}{36} - \frac{(x - 3)^2}{64} = 1 \][/tex]
Thus, the correct equation of the hyperbola is:
[tex]\[ \boxed{\frac{y^2}{36} - \frac{(x - 3)^2}{64} = 1} \][/tex]
1. Determine the center of the hyperbola:
The vertices are located at \((3, -6)\) and \((3, 6)\). The center of the hyperbola is the midpoint of the vertices.
Midpoint [tex]\[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{3 + 3}{2}, \frac{-6 + 6}{2} \right) = (3, 0) \][/tex]
2. Determine the distance between the vertices:
The distance between the two vertices is the length of the transverse axis, which is \(2a\). From \((3, -6)\) to \((3, 6)\), the distance is 12 units.
[tex]\[ 2a = 12 \implies a = \frac{12}{2} = 6 \][/tex]
3. Determine the distance between the foci:
The foci are located at \((3, -10)\) and \((3, 10)\). The distance between the foci is \(2c\). From \((3, -10)\) to \((3, 10)\), the distance is 20 units.
[tex]\[ 2c = 20 \implies c = \frac{20}{2} = 10 \][/tex]
4. Determine \(b\):
Use the relationship between \(a\), \(b\), and \(c\) for hyperbolas:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Substituting the known values:
[tex]\[ 10^2 = 6^2 + b^2 \implies 100 = 36 + b^2 \implies b^2 = 64 \][/tex]
5. Write the equation of the hyperbola:
Since the hyperbola opens vertically (as indicated by the vertical difference in the vertices and foci), its standard form is:
[tex]\[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \][/tex]
Here, \(h = 3\), \(k = 0\), \(a = 6\), and \(b^2 = 64\):
[tex]\[ \frac{(y - 0)^2}{6^2} - \frac{(x - 3)^2}{64} = 1 \implies \frac{y^2}{36} - \frac{(x - 3)^2}{64} = 1 \][/tex]
Thus, the correct equation of the hyperbola is:
[tex]\[ \boxed{\frac{y^2}{36} - \frac{(x - 3)^2}{64} = 1} \][/tex]
