standard form of a hyperbola

Algebra
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write the standard form equation of a hyperbola, with vertices at (8,-3), (-6,-3) and endpoints of conjugate axis at (1,6) and (1, -12)

May 14th, 2015

The standard form equation of this Hyperbola is [(x-1)^2 / 49] - [(y+3)^2 / 81] = 1.


May 14th, 2015

Sure: We have a standard formula for a Hyperbola in which the conjugate axis is vertical, which it is in your case, which is [(x-h)^2 / a^2] - [(y-k)^2 / b^2] = 1. 

Now we must define the constants:

The line joining the two vertices is of length 2a. Here, your line length is 8--6 = 8 + 6 = 14 and so a = 7.

The hyperbola is centered at the point (h, k) which is halfway between your vertices. Therefore in your case, (h, k) = (1, -3): h = 1,  k = -3.

Finally, the length of the conjugate axis is 2b. Since your conjugate axis is 18 units long, b = 9.


Substituting these values into the above equation gives the answer I sent you.


Note: If you have a hyperbola where the conjugate axis is horizontal, the equation above becomes:

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

May 14th, 2015

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May 14th, 2015
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May 14th, 2015
Dec 2nd, 2016
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