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
Shwetank Mauria
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 (+-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]}

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