The equation of hyperbola is [tex]\frac{(y-5)^{2} }{9 } -\frac{(x-3)^{2} }{16 } = 1[/tex]
What is a hyperbola?
A plane curve generated by a point so moving that the difference of the distances from two fixed points is a constant : a curve formed by the intersection of a double right circular cone with a plane that cuts both halves of the cone.
What is standard hyperbola equations?
• If the foci lie on the x-axis, the standard form of a hyperbola can be given as,
[tex]\frac{(x-h)^{2} }{a^{2} } - \frac{(y-k)^{2} }{b^{2} } = 1[/tex]
• If the foci lie on the y-axis, the standard form of the hyperbola is given as,
[tex]\frac{(y-k)^{2} }{a^{2} } -\frac{(x-h)^{2} }{b^{2} } = 1[/tex]
• Coordinates of the center: (h, k).
• Coordinates of vertices: (h+a, k) and (h - a,k)
• Co-vertices correspond to b, the ” minor semi-axis length”, and coordinates of co-vertices: (h,k+b) and (h,k-b).
• Foci have coordinates (h+c,k) and (h-c,k). The value of c is given as, c2 = a2 + b2.
• Slopes of asymptotes: y = ±(b/a)x.
According to the question
Foci = (3,7) (3,-3)
Vertices = (3,5) (3,-1)
As per coordinates and graph the hyperbola is vertical so x will be constant y varies
Distance between foci = 2c (where c is center)
10 = 2c
C = 5
Therefore Centre (h,k) = (3,5)
Distance between vertices = 2a
2a= 6
a = 3
The value of c is
[tex]c^{2} = a^{2} + b^{2}[/tex]
25 = 9 + [tex]b^{2}[/tex]
[tex]b^{2}[/tex] = 16
Now as per our equation
h=3 , k=5 , [tex]a^{2}[/tex] = 9, [tex]b^{2}[/tex] = 16
[tex]\frac{(y-k)^{2} }{a^{2} } -\frac{(x-h)^{2} }{b^{2} } = 1[/tex]
[tex]\frac{(y-5)^{2} }{9 } -\frac{(x-3)^{2} }{16 } = 1[/tex]
Hence, equation of hyperbola = [tex]\frac{(y-5)^{2} }{9 } -\frac{(x-3)^{2} }{16 } = 1[/tex]
To know more about hyperbola here :
https://brainly.com/question/15697124
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