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

Aug 4, 2017

${x}^{2} / 9 - {y}^{2} / 16 = 1$

#### 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 $\left(\pm 3 , 0\right)$ and focii $\left(\pm 5 , 0\right)$, 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 $b = \pm 4$

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]}