SOLUTION: what is the standard form of the equation of a parabola with the general form of x^2-9y^2+4x-36y-41=0?

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Question 1090308: what is the standard form of the equation of a parabola with the general form of x^2-9y^2+4x-36y-41=0?
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52809) About Me  (Show Source):
You can put this solution on YOUR website!
.
Parabola ??

It is not an equation of a parabola.

It is the equation of a hyperbola, instead.



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-9y%5E2%2B4x-36y-41=0 => this is a hyperbola

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

rearrange terms and complete squares:

x%5E2%2B4x-9y%5E2-36y-41=0

%28x%5E2%2B4x%2Bb%5E2%29-b%5E2-9%28y%5E2%2B4y%2Bb%5E2%29-9b%5E2=41

%28x%5E2%2B4x%2B2%5E2%29-2%5E2-9%28y%5E2%2B4y%2B2%5E2%29-9%2A2%5E2=41

%28x%2B2%29%5E2-4-9%28y%2B2%29%5E2-9%2A4=41

%28x%2B2%29%5E2-9%28y%2B2%29%5E2-36-4=41

%28x%2B2%29%5E2-9%28y%2B2%29%5E2-40=41

%28x%2B2%29%5E2-9%28y%2B2%29%5E2=41%2B40

%28x%2B2%29%5E2-9%28y%2B2%29%5E2=81

%28x%2B2%29%5E2%2F81-9%28y%2B2%29%5E2%2F81=81%2F81

%28x%2B2%29%5E2%2F81-cross%289%29%28y%2B2%29%5E2%2Fcross%2881%299=1

%28x%2B2%29%5E2%2F81-%28y%2B2%29%5E2%2F9=1

%28x%2B2%29%5E2%2F9%5E2-%28y%2B2%29%5E2%2F3%5E2=1
=>h=-2, k=-2,a=9, b=3
center at (-2,-2)