The answer is:
[tex] \frac{ y^{2} }{256} } - \frac{ x^{2} }{ 144 } = 1[/tex]
Explanation:
The general equation for a hyperbola with transverse axis vertical is
[tex] \frac{ (y - k)^{2} }{ a^{2} } - \frac{ (x - h)^{2} }{ b^{2} } = 1[/tex]
where:
h = x-coordinate of the center
k = y-coordinate of the center
a = semi-major axis
b = semi-minor axis
We know that the center has coordinates (0, 0), therefore:
h = 0
k = 0
We also know that
a = 16
2c = distance between foci = 40
In order to find b², we need to use the formula
c² = a² + b²
and solve it for b²:
b² = c² - a²
= (40÷2)² - 16²
= 400 - 256
= 144
Now we can substitute the obtained values in the general equation to find the required equation:
[tex] \frac{ y^{2} }{256} } - \frac{ x^{2} }{ 144 } = 1[/tex]
