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

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

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Answer 1 · 2017-08-04T16:33:21

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

Explanation:

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

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ $x^2/a^2-y^2/b^2=1$

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

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

Hence equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$ $x^2/9-y^2/16=1$

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]}

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

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