How do you write the equation of a hyperbola given center at the origin (0,0) and satisfies the given conditions: Foci F(0, -5) (0,5) and conjugate axis of length 4?

1 Answer
Jul 23, 2018

Answer:

The equation is #x^2/21-y^2/4=1#

Explanation:

The general equation of a hyperbola is

#(x-h)^2/a^2-(y-k)^2/b^2=1#

The center is #C=(h,k)=(0,0)#

The equation reduces to

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

The conjugate axis is

#2b=4#, #=>#, #b=2#

The equation is

#x^2/a^2-y^2/4=1#

The foci are #F=(0,c)=(0,5)# and #F'=(0,-c)=(0,-5)#

And

#c^2=a^2+b^2#

Therefore,

#a^2=c^2-b^2=5^2-4=21#

Therefore,

The equation is

#x^2/21-y^2/4=1#

graph{x^2/21-y^2/4=1 [-18.02, 18.02, -9.01, 9.02]}