How do you find an equation of the hyperbola with its center at the origin. Vertices (+OR- 3,0); Foci: (+OR- 5,0)?

1 Answer
Aug 4, 2017

Answer:

#x^2/9-y^2/16=1#

Explanation:

The equation of the hyperbola with its center at the origin is

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

As ordinate is common in vertices #(+-3,0)# and focii #(+-5,0)#, it is a horizontal parabolan further #a=3#.

Further distance from center to either focii is #5#, hence #c=5#, where #c^2=a^2+b^2# and hence #b^2=25-9=16# and #b=+-4#

Hence equation of hyperbola is #x^2/9-y^2/16=1#

graph{(x^2/9-y^2/16-1)((x-5)^2+y^2-0.01)((x+5)^2+y^2-0.01)((x-3)^2+y^2-0.01)(x^2+y^2-0.01)((x+3)^2+y^2-0.01)=0 [-10, 10, -5, 5]}