SOLUTION: Transform the equation of the hyperbola from standard from into general form (y^2)/36 - (x^2)/16 = 1 (9x^2) - (4y^2) + 144 = 0 is the answer But I keep getting (-9x^2) +

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Transform the equation of the hyperbola from standard from into general form (y^2)/36 - (x^2)/16 = 1 (9x^2) - (4y^2) + 144 = 0 is the answer But I keep getting (-9x^2) +      Log On


   

Question 1204696: Transform the equation of the hyperbola from standard from into general form
(y^2)/36 - (x^2)/16 = 1
(9x^2) - (4y^2) + 144 = 0 is the answer
But I keep getting (-9x^2) + (4y^2) - 144 = 0
Why are all my positive and negative signs reversed?
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
The general form of a hyperbola is
Ax%5E2+%2B+Bx+%2B+Cy%5E2+%2B+Dy+%2B+E+=+0
where either A or C+is negative (but never both)
respecting alphabetic order, let say A is positive and C is negative
Ax%5E2+%2B+Bx+-Cy%5E2+%2B+Dy+%2B+E+=+0
in this case hyperbola opens up and down, just as in your case
%28y%5E2%29%2F36+-+%28x%5E2%29%2F16+=+1....both sides multiply by 36%2A16 and you get
16y%5E2-+36x%5E2+=+36%2A16... simplify, divide by 4
4y%5E2-+9x%5E2+=+36%2A4
4y%5E2-+9x%5E2+-144=0.... A=-9, to make it positive multiply equation by -1
-4y%5E2%2B9x%5E2+%2B144=0...rearrange
9x%5E2+-4y%5E2%2B144=0



Answer by ikleyn(52814) About Me  (Show Source):
You can put this solution on YOUR website!
.

Your equation in the post is mathematically equivalent to the "answer" equation.


One equation is precisely another multiplied by -1.

Multiplication by a constant non-zero number leaves/makes/remains equations equivalent.

So, you may not trouble.