Answer:
(y+3)^2/8^2 - (x-1)^2/6^2 = 1
Step-by-step explanation:
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The equation of the hyperbola is (y+3)^2/64 - (x-1)^2/36 = 1
It is a curve in two-dimensional geometry having two parts they both are symmetric. In other words, It can be defined as the number of points in the two-dimensional geometry that have a constant difference from that point to two fixed points in the plane.
We have foci of the hyperbola at (1, 7) and (1, -13)
We know the foci of the hyperbola are (α, β+ae) and (α, β-ae)
Where (α, β) is the center of the hyperbola.
Compare the above foci with standard foci, we get:
β+ae = 7 ..... (1)
β-ae = -13 .....(2)
Add the above two equations, we get:
2β = -6
β = -3
Now, subtract the above two equations, we get:
2ae = 20 ....(3)
We have directrix of the hyperbola:
y = 64/10
The standard equation of the hyperbola is y = a/e after comparing, we get:
[tex]\rm \frac{a}{e} = \frac{64}{10}[/tex] ....(4)
solve the equation (3) and (4), we get:
a = 8 and e = 10/8
We know the [tex]\rm b^2 = a^2(e^2-1)[/tex]
[tex]\rm b^2 = 8^2(\frac{10}{8} ^2-1)[/tex]
b = 6
The standard equation of the hyperbola is given by:
[tex]\rm \frac{(y-\beta)^2}{a^2} - \frac{(x-\alpha)^2}{b^2} = 1[/tex]
[tex]\rm \frac{(y+3)^2}{8^2} - \frac{(x-1)^2}{6^2} = 1[/tex] ( α = 1, β = -3, a = 8, and b = 6)
[tex]\rm \frac{(y+3)^2}{64} - \frac{(x-1)^2}{36} = 1[/tex]
Thus, the equation of the hyperbola is (y+3)^2/64 - (x-1)^2/36 = 1
Learn more about the hyperbola here:
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