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Question 1120388: what type of conic is this equation x^2-2y^2+2x+8y-7=0? how can we get the center, the a and the b?
Found 2 solutions by solver91311, ikleyn: Answer by solver91311(24713) (Show Source): Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
x^2 - 2y^2 + 2x + 8y - 7 = 0.
Complete the squares:
(x^2 + 2x) - (2y^2 - 8y) = 7
(x^2 + 2x + 1) - 2*(y^2 - 4y + 4) = 7 + 1 - 8
(x+1)^2 - 2*(y-2)^2 = 0. (*)
IT IS NOT a PARABOLA; it is not a CIRCLE; it is not an ELLIPSE; finally, it is not a HYPERBOLA.
Then, what is it ?
It is an equation of two straight lines.
How to see (how to prove) it ? -- For it, continue with the equation (*)
You see the difference of the two squares in its left side, isn't it ?
Hence, you can rewrite it in this EQUIVALENT form as the product of the sum and the difference of its terms
(after taking square roots from them):
= 0, or, equivalently
= 0. (*)
Thus it is the product of the two linear functions
f(x,y) =
and
g(x,y) = .
Therefore, equation (*) represents the UNION of points, belonging to one of the two straight lines
= 0 (1)
and/or
= 0. (2)
These lines are shown in the figure below: line (1) in red and line (2) in green.
Plot = 0 (red) and plot = 0 (green)
It is a quite rare case when a hyperbola DEGENERATES into the union of its two straight line asymptotes.
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