A hyperbola has the vertices $(0,0)$ and $(0,-16)$ and the foci $(0,2)$ and $(0,-18)$. Find the equation with the given information.
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Please show what form you assumed and work done one so far. – Narasimham Apr 08 '16 at 19:43
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So far the work I've been able to gather is [(x^2)/a^2]-[(y+16)^2/b^2] I'm unsure on how to find the value for "a" or "b" – Mike Apr 08 '16 at 19:47
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You are not off to a good start. – Doug M Apr 08 '16 at 19:50
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I agree. It's frustrating. The text we have for class doesn't state how to find an equation from information as such. – Mike Apr 08 '16 at 19:51
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Where would be the fun in plugging numbers into an equation that the text has provided for you? – Doug M Apr 08 '16 at 19:54
2 Answers
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Your vertices and foci lie on the y axis. This means that your hyperbola opens upward.
Equation for a generic hyperbola that opens upward:
$-\frac{(x-h)^2}{a^2} + \frac{(y-v)^2}{b^2} = 1$
The center $(h,v)$ lies at the mid-point between the two foci.
$a$ = distance from center to each vertex.
$c^2 = a^2+b^2$, Where c is the distance from the center to the focus.
Doug M
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There is enough information given to find a, and c. And based on what I have written what do you think that says about b? – Doug M Apr 08 '16 at 19:51
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I think I've got it! I found b to equal 6. I think I can take it from here! – Mike Apr 08 '16 at 19:56
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The hyperbola is opening up in direction $y:$
$$ a = 8;\, b = 6; c = \sqrt {a^2 + b^2 } = a\,e =10\,;$$
Carefully choose sign to shift axis downward so the top vertex touches x-axis.
$$ \dfrac{(y+a)^2}{a^2} - \dfrac {x^2}{b^2} =1. $$
Narasimham
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