How do you write a standard form equation for the hyperbola with #36x^2 - 100y^2 - 72x + 400y = 3964#?

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Answer 1 · 2016-03-06T07:13:48

#[(x - 3/2)^2]/((47sqrt3)/6)^2 - [(y - 2)^2]/((47sqrt2)/10)^2 = 1#

The standard form looks something like this

#frac{(x-h)^2}{a^2} - frac{(y-k)^2}{b^2} = 1#,

where #h#, #k#, #a# and #b# are constants to be determined.

First thing to do is to simplify the equation. Divide everything by 4.

#6x^2 - 25y^2 - 18x + 100y = 991#

Next, factorize.

#6(x^2 - 3x) - 25(y^2 - 4y) = 991#

Next, complete the square

#6(x - 3/2)^2 - 25(y - 2)^2 = 991 + 6(3/2)^2 + 25(2)^2#

#= 2209/2#

Divide both sides by 2209/2 to get 1 on the right hand side.

#[(x - 3/2)^2]/(2209/12) - [(y - 2)^2]/(2209/50) = 1#

Take the square root of the denominator, and square it afterwards to make it into standard form.

#[(x - 3/2)^2]/((47sqrt3)/6)^2 - [(y - 2)^2]/((47sqrt2)/10)^2 = 1#