Figuring out Standard equations of Hyperbola

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gopher

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Jan 14, 2007
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ok so im trying to put an equation into the stand form for a hyperbola
heres is what i figured out so far tell me where i am doing wrong

4x^2-9y^2-32x+36y+27=0
4x^2-32x-9y^2+36y=-27 #rearrange
4(x^2-8x)-9(y^2-4x)=-27 #factor
4(x^2-8x+16)-9(y^2-4y+4)=-27+4(16)-9(4) #completing the square
4(x-4)^2-9(y+2)^2=1 #factor and simplify

Is that correct so far ?
If so how do i get that into the form???

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

thanks
 
Yes it is correct and \(\displaystyle \frac{{\left( {x - 4} \right)^2 }}{{\frac{1}{4}}} - \frac{{\left( {y + 2} \right)^2 }}{{\frac{1}{9}}} = 1.\)
 
cool thanks its good to know im heading in the right direction
I hit another stumbling block
given:
4x^2-3y^2-32x+6y+73=0
4x^2-32x-3y^2+6y=-73
4(x^2-8x)-3(y^2+2y)=-73
4(x^2-8x+16)-3(y^2+2y+1)=-73+4(16)-3(1)
4(x-4)^2-3(y+1)^2=-12
i get
(((x-4)^2)/-3) - (((y-1)^2)/-4) = 1
which obviously is not the correct answer
in the back of the book the correct answer is
(((y-1)^2)/4) - (((x-4)^2)/3) = 1
any ideas what i did wrong
thanks again
gopher
 
Your division is wrong!
Divide every term by -12.
The you get the text's answer.
 
Hello Gopher,

One minor correction on the first problem, I believe -- be careful with your signs:

(y^2-4y+4) = (y-2)^2
 
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