The equation of the hyperbola is [tex]\frac{(y + 3)^2}{64} -\frac{(x - 1)^2}{36} = 1[/tex]
The foci are given as:
(1, 7) and (1, -13)
The foci of a hyperbola are represented using: (m, n+p) and (m, n-p)
Where the center of the hyperbola is (n,p)
By comparison, we have:
m = 1
n + p = 7
n - p =13
Where n = ae
The solution to the above system of equations is
m = 1, n = 10 and p = -3
The directrix is given as:
y = 64/10
The directrix is represented as:
y = a/e
So, we have:
a/e = 64/10
Substitute n = ae in n = 10
ae = 10
Solving ae = 10 and a/e = 64/10, we have:
a = 8 and e = 10/8
To calculate b, we make use of:
[tex]b = a\sqrt{(e^2 - 1)[/tex]
So, we have:
[tex]b = 8\sqrt{(\frac{10}{8}^2 - 1)[/tex]
Evaluate
b = 8 * 0.75
b = 6
The equation of the hyperbola is then calculated using:
[tex]\frac{(y - p)^2}{a^2} -\frac{(x - m)^2}{b^2} = 1[/tex]
This gives
[tex]\frac{(y + 3)^2}{8^2} -\frac{(x - 1)^2}{6^2} = 1[/tex]
Evaluate
[tex]\frac{(y + 3)^2}{64} -\frac{(x - 1)^2}{36} = 1[/tex]
Hence, the equation of the hyperbola is [tex]\frac{(y + 3)^2}{64} -\frac{(x - 1)^2}{36} = 1[/tex]
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