Answer :
To determine which hyperbola has both foci lying in the same quadrant, we'll need to analyze the standard forms of the hyperbola equations given and find the coordinates of their centers.
The standard form of a hyperbola equation can be either:
1. \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) for a hyperbola opening left-right (horizontal hyperbola)
2. \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) for a hyperbola opening up-down (vertical hyperbola)
In both forms, the center of the hyperbola is \((h, k)\).
Let's analyze each option:
### Option A
[tex]\[ \frac{(x-24)^2}{24^2} - \frac{(y-1)^2}{7^2} = 1 \][/tex]
This is a horizontal hyperbola with its center at \((24, 1)\). The foci for a horizontal hyperbola are given by:
\((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2}\).
### Option B
[tex]\[ \frac{(y-12)^2}{5^2} - \frac{(x-6)^2}{12^2} = 1 \][/tex]
This is a vertical hyperbola with its center at \((6, 12)\). The foci for a vertical hyperbola are given by:
\((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
### Option C
[tex]\[ \frac{(y-16)^2}{15^2} - \frac{(x-2)^2}{8^2} = 1 \][/tex]
This is a vertical hyperbola with its center at \((2, 16)\). The foci for a vertical hyperbola are given by:
\((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
### Option D
[tex]\[ \frac{(y-16)^2}{9^2} - \frac{(x+1)^2}{12^2} = 1 \][/tex]
This is a vertical hyperbola with its center at \((-1, 16)\). The foci for a vertical hyperbola are given by:
\((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
Now, calculate \( c \) for each hyperbola:
- Option A:
[tex]\[ c = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \][/tex]
Thus, the foci are at \((24 \pm 25, 1)\), which are at \((-1, 1)\) and \((49, 1)\).
- Option B:
[tex]\[ c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \][/tex]
Thus, the foci are at \((6, 12 \pm 13)\), which are at \((6, -1)\) and \((6, 25)\).
- Option C:
[tex]\[ c = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \][/tex]
Thus, the foci are at \((2, 16 \pm 17)\), which are at \((2, -1)\) and \((2, 33)\).
- Option D:
[tex]\[ c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \][/tex]
Thus, the foci are at \((-1, 16 \pm 15)\), which are at \((-1, 1)\) and \((-1, 31)\).
To lie in the same quadrant, both coordinates of the foci must have the same sign in each axis. Checking the above calculations:
- Option A: Foci \((-1, 1)\) and \((49, 1)\) - These are not in the same quadrant.
- Option B: Foci \((6, -1)\) and \((6, 25)\) - These are not in the same quadrant.
- Option C: Foci \((2, -1)\) and \((2, 33)\) - These are not in the same quadrant.
- Option D: Foci \((-1, 1)\) and \((-1, 31)\) - These are in the same quadrant.
Thus, the correct answer is:
[tex]\[ \boxed{D} \][/tex]
The standard form of a hyperbola equation can be either:
1. \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) for a hyperbola opening left-right (horizontal hyperbola)
2. \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) for a hyperbola opening up-down (vertical hyperbola)
In both forms, the center of the hyperbola is \((h, k)\).
Let's analyze each option:
### Option A
[tex]\[ \frac{(x-24)^2}{24^2} - \frac{(y-1)^2}{7^2} = 1 \][/tex]
This is a horizontal hyperbola with its center at \((24, 1)\). The foci for a horizontal hyperbola are given by:
\((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2}\).
### Option B
[tex]\[ \frac{(y-12)^2}{5^2} - \frac{(x-6)^2}{12^2} = 1 \][/tex]
This is a vertical hyperbola with its center at \((6, 12)\). The foci for a vertical hyperbola are given by:
\((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
### Option C
[tex]\[ \frac{(y-16)^2}{15^2} - \frac{(x-2)^2}{8^2} = 1 \][/tex]
This is a vertical hyperbola with its center at \((2, 16)\). The foci for a vertical hyperbola are given by:
\((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
### Option D
[tex]\[ \frac{(y-16)^2}{9^2} - \frac{(x+1)^2}{12^2} = 1 \][/tex]
This is a vertical hyperbola with its center at \((-1, 16)\). The foci for a vertical hyperbola are given by:
\((h, k \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
Now, calculate \( c \) for each hyperbola:
- Option A:
[tex]\[ c = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \][/tex]
Thus, the foci are at \((24 \pm 25, 1)\), which are at \((-1, 1)\) and \((49, 1)\).
- Option B:
[tex]\[ c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \][/tex]
Thus, the foci are at \((6, 12 \pm 13)\), which are at \((6, -1)\) and \((6, 25)\).
- Option C:
[tex]\[ c = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \][/tex]
Thus, the foci are at \((2, 16 \pm 17)\), which are at \((2, -1)\) and \((2, 33)\).
- Option D:
[tex]\[ c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \][/tex]
Thus, the foci are at \((-1, 16 \pm 15)\), which are at \((-1, 1)\) and \((-1, 31)\).
To lie in the same quadrant, both coordinates of the foci must have the same sign in each axis. Checking the above calculations:
- Option A: Foci \((-1, 1)\) and \((49, 1)\) - These are not in the same quadrant.
- Option B: Foci \((6, -1)\) and \((6, 25)\) - These are not in the same quadrant.
- Option C: Foci \((2, -1)\) and \((2, 33)\) - These are not in the same quadrant.
- Option D: Foci \((-1, 1)\) and \((-1, 31)\) - These are in the same quadrant.
Thus, the correct answer is:
[tex]\[ \boxed{D} \][/tex]
