Question 448928
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Hi,
Converting each equation to the standard form of conic section
y^2+6y-x=0   |Parabola opening right or left
 x = y^2+6y
 x = (y+3)^2 - 9   |Vertex form V(-9,-3)
 
x^2-y^2-6x+13=0  |Hyperbola opening up and down (multiplying thru by -1)
 y^2 - x^2 + 6x -13 = 0
 y^2 -[(x-3)^2 -9] - 13 = 0
 y^2 -(x-3)^2 +9 - 13 = 0
 y^2 - (x-3)^2 = 4 
{{{y^2/4 - (x-3)^2/4 = 1}}}