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To find the standard form of the equation of a hyperbola that has vertices at [tex]\((-10, -15)\)[/tex] and [tex]\((70, -15)\)[/tex] and one of its foci at [tex]\((-11, -15)\)[/tex], we need to follow these steps:
1. Determine the center of the hyperbola:
The center of the hyperbola is the midpoint of the vertices.
- For the vertices [tex]\((-10, -15)\)[/tex] and [tex]\((70, -15)\)[/tex]:
[tex]\[ \text{Center} = \left( \frac{-10 + 70}{2}, -15 \right) = (30, -15) \][/tex]
2. Find the distance [tex]\(2a\)[/tex] between the vertices:
The distance between the vertices is [tex]\(2a\)[/tex].
- For the vertices [tex]\((-10, -15)\)[/tex] and [tex]\((70, -15)\)[/tex]:
[tex]\[ 2a = 70 - (-10) = 70 + 10 = 80 \quad \Rightarrow \quad a = \frac{80}{2} = 40 \][/tex]
3. Determine the distance [tex]\(c\)[/tex] to the foci:
The distance from the center to each focus is [tex]\(c\)[/tex].
- Since one of the foci is at [tex]\((-11, -15)\)[/tex] and the center is at [tex]\((30, -15)\)[/tex]:
[tex]\[ c = 30 - (-11) = 30 + 11 = 41 \][/tex]
4. Use the relationship [tex]\(c^2 = a^2 + b^2\)[/tex] to determine [tex]\(b^2\)[/tex]:
[tex]\[ c^2 = a^2 + b^2 \quad \Rightarrow \quad 41^2 = 40^2 + b^2 \quad \Rightarrow \quad 1681 = 1600 + b^2 \quad \Rightarrow \quad b^2 = 1681 - 1600 = 81 \quad \Rightarrow \quad b = \sqrt{81} = 9 \][/tex]
5. Form the standard equation of a hyperbola:
Since the hyperbola opens horizontally (the x-coordinates of the vertices differ), the equation is:
[tex]\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \][/tex]
Here, [tex]\(h = 30\)[/tex], [tex]\(k = -15\)[/tex], [tex]\(a = 40\)[/tex], and [tex]\(b = 9\)[/tex]:
[tex]\[ \frac{(x - 30)^2}{40^2} - \frac{(y + 15)^2}{9^2} = 1 \][/tex]
Putting it all together:
[tex]\[ \frac{(x - 30)^2}{1600} - \frac{(y + 15)^2}{81} = 1 \][/tex]
So the complete standard form of the equation of the hyperbola is:
[tex]\[ \frac{(x - 30)^2}{1600} - \frac{(y + 15)^2}{81} = 1 \][/tex]
1. Determine the center of the hyperbola:
The center of the hyperbola is the midpoint of the vertices.
- For the vertices [tex]\((-10, -15)\)[/tex] and [tex]\((70, -15)\)[/tex]:
[tex]\[ \text{Center} = \left( \frac{-10 + 70}{2}, -15 \right) = (30, -15) \][/tex]
2. Find the distance [tex]\(2a\)[/tex] between the vertices:
The distance between the vertices is [tex]\(2a\)[/tex].
- For the vertices [tex]\((-10, -15)\)[/tex] and [tex]\((70, -15)\)[/tex]:
[tex]\[ 2a = 70 - (-10) = 70 + 10 = 80 \quad \Rightarrow \quad a = \frac{80}{2} = 40 \][/tex]
3. Determine the distance [tex]\(c\)[/tex] to the foci:
The distance from the center to each focus is [tex]\(c\)[/tex].
- Since one of the foci is at [tex]\((-11, -15)\)[/tex] and the center is at [tex]\((30, -15)\)[/tex]:
[tex]\[ c = 30 - (-11) = 30 + 11 = 41 \][/tex]
4. Use the relationship [tex]\(c^2 = a^2 + b^2\)[/tex] to determine [tex]\(b^2\)[/tex]:
[tex]\[ c^2 = a^2 + b^2 \quad \Rightarrow \quad 41^2 = 40^2 + b^2 \quad \Rightarrow \quad 1681 = 1600 + b^2 \quad \Rightarrow \quad b^2 = 1681 - 1600 = 81 \quad \Rightarrow \quad b = \sqrt{81} = 9 \][/tex]
5. Form the standard equation of a hyperbola:
Since the hyperbola opens horizontally (the x-coordinates of the vertices differ), the equation is:
[tex]\[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \][/tex]
Here, [tex]\(h = 30\)[/tex], [tex]\(k = -15\)[/tex], [tex]\(a = 40\)[/tex], and [tex]\(b = 9\)[/tex]:
[tex]\[ \frac{(x - 30)^2}{40^2} - \frac{(y + 15)^2}{9^2} = 1 \][/tex]
Putting it all together:
[tex]\[ \frac{(x - 30)^2}{1600} - \frac{(y + 15)^2}{81} = 1 \][/tex]
So the complete standard form of the equation of the hyperbola is:
[tex]\[ \frac{(x - 30)^2}{1600} - \frac{(y + 15)^2}{81} = 1 \][/tex]
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