Question

Find an equation in standard form for the hyperbola with vertices at (0, ±4) and foci at (0, ±5).

y squared over 16 minus x squared over 9 = 1
y squared over 25 minus x squared over 16 = 1
y squared over 16 minus x squared over 25 = 1
y squared over 9 minus x squared over 16 = 1

Answer 1

please help!!

Answer 2

@Shimansk123 why choice b. is correct ? please explain your answer !!!

Answer 3

I haven't done this in a while but I hope I can explain this well:

since the vertices and foci are equidistant from the origin & lie on the x-axis, it is reasonable to conclude that the hyperbola is centered on the origin, which means our h and k values are zero. therefore we only need to focus on finding our a, b, and c values.

x^2/a^2 - y^2/b^2 = 1

where a is the distance between the center and vertex (the distance between (0,0) and (0,4), or (0,0) and (0,-4))
and c is the distance between the center and the focii (the distance between (0,0) and (0,5), or (0,0) and (0,-5))

from there, b^2 = c^2 - a^2

once you have a and b you can plug them into the hyperbola equation