Question

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

Answer 1

`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]}