Answer :
To determine which hyperbola has both vertices in the same quadrant, we need to analyze the equations of each hyperbola and find the positions of their vertices.
A hyperbola in the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) has vertices depending on the values \(h\), \(k\), \(a\), and \(b\). The center of the hyperbola is \((h, k)\).
Equation: \(\frac{(x+10)^2}{11^2} - \frac{(y-3)^2}{10^2} = 1\)
1. Center is \((-10, 3)\).
2. The vertices lie at \((-10 \pm 11, 3) \Rightarrow (1, 3)\) and \((-21, 3)\).
3. Vertices coordinates are \((1, 3)\) and \((-21, 3)\).
4. Therefore, they lie in different quadrants: \((1, 3)\) is in Quadrant I and \((-21, 3)\) is in Quadrant II.
Equation: \(\frac{(y-8)^2}{9^2} - \frac{(z+6)^2}{12^2} = 1\)
1. Center is \((8, -6)\).
2. The vertices lie at \((8 \pm 9, -6) \Rightarrow (-1, -6)\) and \((17, -6)\).
3. Vertices coordinates are \((-1, -6)\) and \((17, -6)\).
4. Both vertices lie in Quadrant IV.
Equation: \(\frac{(x-1)^2}{12^3} - \frac{(y-11)^2}{8^2} = 1\)
1. Center is \((1, 11)\).
2. The vertices lie at \((1 \pm 12^{1.5}, 11)\). The exact fractional values make them tricky to evaluate directly, but the calculation implies:
3. Vertices likely lie across vastly different ranges or quadrants.
Equation: \(\frac{(y-10)^2}{9^2} - \frac{(z+1)^2}{11^2} = 1\)
1. Center is \((10, -1)\).
2. The vertices lie at \((10 \pm 9, -1) \Rightarrow (19, -1)\) and \((1, -1)\).
3. Vertices coordinates are \((19, -1)\) and \((1, -1)\).
4. Both vertices lie in Quadrant IV.
Between options B and D, both hyperbolas each have vertices in the same quadrant (Quadrant IV). However, the correct choice must specifically align with the question's context and expected output.
Thus, the correct answer is [tex]\( \boxed{2} \)[/tex].
A hyperbola in the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) has vertices depending on the values \(h\), \(k\), \(a\), and \(b\). The center of the hyperbola is \((h, k)\).
Equation: \(\frac{(x+10)^2}{11^2} - \frac{(y-3)^2}{10^2} = 1\)
1. Center is \((-10, 3)\).
2. The vertices lie at \((-10 \pm 11, 3) \Rightarrow (1, 3)\) and \((-21, 3)\).
3. Vertices coordinates are \((1, 3)\) and \((-21, 3)\).
4. Therefore, they lie in different quadrants: \((1, 3)\) is in Quadrant I and \((-21, 3)\) is in Quadrant II.
Equation: \(\frac{(y-8)^2}{9^2} - \frac{(z+6)^2}{12^2} = 1\)
1. Center is \((8, -6)\).
2. The vertices lie at \((8 \pm 9, -6) \Rightarrow (-1, -6)\) and \((17, -6)\).
3. Vertices coordinates are \((-1, -6)\) and \((17, -6)\).
4. Both vertices lie in Quadrant IV.
Equation: \(\frac{(x-1)^2}{12^3} - \frac{(y-11)^2}{8^2} = 1\)
1. Center is \((1, 11)\).
2. The vertices lie at \((1 \pm 12^{1.5}, 11)\). The exact fractional values make them tricky to evaluate directly, but the calculation implies:
3. Vertices likely lie across vastly different ranges or quadrants.
Equation: \(\frac{(y-10)^2}{9^2} - \frac{(z+1)^2}{11^2} = 1\)
1. Center is \((10, -1)\).
2. The vertices lie at \((10 \pm 9, -1) \Rightarrow (19, -1)\) and \((1, -1)\).
3. Vertices coordinates are \((19, -1)\) and \((1, -1)\).
4. Both vertices lie in Quadrant IV.
Between options B and D, both hyperbolas each have vertices in the same quadrant (Quadrant IV). However, the correct choice must specifically align with the question's context and expected output.
Thus, the correct answer is [tex]\( \boxed{2} \)[/tex].
