Question 1120388
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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 <U>NOT a  PARABOLA</U>;  it is <U>not a CIRCLE</U>;  it is <U>not an ELLIPSE</U>;   finally, it is <U>not a HYPERBOLA</U>.


    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):


{{{(x+1 - sqrt(2)*(y-2)) * (x+1 + sqrt(2)*(y-2))}}} = 0,     or, equivalently


{{{(x - sqrt(2)*y + (1+2*sqrt(2))) * (x + sqrt(2)*y + (1-2*sqrt(2)))}}} = 0.    (*)


Thus it is the product of the two linear functions

    f(x,y) = {{{x - sqrt(2)*y + (1+2*sqrt(2))}}} 

and

    g(x,y) = {{{x + sqrt(2)*y + (1-2*sqrt(2))}}}.


Therefore, equation  (*)  represents the UNION  of points, belonging to one of the two straight lines

    {{{x - sqrt(2)*y + (1+2*sqrt(2))}}} = 0      (1)

and/or

    {{{x + sqrt(2)*y + (1-2*sqrt(2))}}} = 0.     (2)


These lines are shown in the figure below: line (1) in red and line (2) in green.



{{{graph( 330, 330, -5.5, 5.5, -5.5, 5.5,
          (x + (1+2*sqrt(2)))/sqrt(2),  (-x - (1-2*sqrt(2)))/sqrt(2)
)}}}


Plot  {{{x - sqrt(2)*y + (1+2*sqrt(2))}}} = 0 (red) and plot  {{{x + sqrt(2)*y + (1-2*sqrt(2))}}} = 0 (green)
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It is a quite rare case when a hyperbola &nbsp;<U>DEGENERATES</U>&nbsp; into the union of its two straight line asymptotes.