Complete 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][tex]$(-11,-15)$[/tex][/tex].

Drag each sign and value to the correct location on the equation. Each sign and value can be used more than once, but not all signs and values will be used.

Options:
- 30
- 15
- 21
- 7
- 40

[tex]\square[/tex]

Reset



Answer :

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]