How do you write the equation of a hyperbola given center at the origin (0,0) and satisfies the given conditions: Foci F(0, -5) (0,5) and conjugate axis of length 4?

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Answer 1 · 2018-07-23T07:20:14

The equation is $x^2/21-y^2/4=1$

The general equation of a hyperbola is

$(x-h)^2/a^2-(y-k)^2/b^2=1$

The center is $C=(h,k)=(0,0)$

The equation reduces to

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

The conjugate axis is

$2b=4$ , $=>$ , $b=2$

The equation is

$x^2/a^2-y^2/4=1$

The foci are $F=(0,c)=(0,5)$ and $F'=(0,-c)=(0,-5)$

And

$c^2=a^2+b^2$

Therefore,

$a^2=c^2-b^2=5^2-4=21$

Therefore,

The equation is

$x^2/21-y^2/4=1$

graph{x^2/21-y^2/4=1 [-18.02, 18.02, -9.01, 9.02]}