Respuesta :
Answer:
The equation is [tex]\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1[/tex].
Step-by-step explanation:
The equation of a hyperbola is represented by [tex]\frac{(y - k)^{2} }{b^{2} } - \frac{(x - h)^{2} }{a^{2} } = 1[/tex], where (h, k) is the center of the hyperbola.
As per the given condition, h = -2 and k = 4.
Thus, the equation becomes [tex]\frac{(y - 4)^{2} }{b^{2} } - \frac{(x + 2)^{2} }{a^{2} } = 1[/tex].
One vertex is (-2, 7).
Hence, putting x = -2 and y = 7, we get [tex]b^{2} = 9[/tex].
The slope of the asymptote is [tex]\frac{b}{a}[/tex].
Hence, [tex]\frac{b^{2} }{a^{2} } = \frac{1}{4} \\a^{2} = 36[/tex].
Thus, the equation is [tex]\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1[/tex].
The equation of the hyperbola in standard form is [tex]\frac{(y-7)^2}{9} - \frac{(x+2))^2}{36}= 1[/tex]
The vertex of the hyperbola, (h, k) = (-2, 7)
The center of a hyperbola, (a, b) = (-2, 4)
The equation of a hyperbola is given as:
[tex]\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2}= 1[/tex]
Substitute h = -2, k = 7, a = -2, b = 4 into the equation [tex]\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2}= 1[/tex]
[tex]\frac{(4-7)^2}{b^2} - \frac{(-2-(-2))^2}{a^2}= 1\\\\\frac{(-3)^2}{b^2} =1\\\\\frac{9}{b^2} = 1\\\\b^2=9\\\\b=\sqrt{9} \\\\b = 3[/tex]
The asymptote of the hyperbola = b/a
b/a = 1/2
3/a = 1/2
a = 6
Substitute a = 6, b = 3, h = -2, k = 7 into the equation [tex]\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2}= 1[/tex]
[tex]\frac{(y-7)^2}{4^2} - \frac{(x-(-2))^2}{3^2}= 1\\\\\frac{(y-7)^2}{9} - \frac{(x+2))^2}{36}= 1[/tex]
The equation of the hyperbola is:
[tex]\frac{(y-7)^2}{9} - \frac{(x+2))^2}{36}= 1[/tex]
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