Asked by joey
fin the center ,principal axis vertices, sketch the graph of 4x^-y^2 +56 x +2y+195=0
Answers
Answered by
Reiny
let's complete the square ...
4x^2 + 56x - (y^2 - 2y ) = -195
4(x^2 + 14x + <b>49</b>) - (y^2 - 2y + <b>1</b>) = -195 + 4(49) - 1
4(x+7)^2 - (y-1)^2 = 0
ahh, we have a degenerated hyperbola
the hyperbola has degenerated into its asymtotes
notice we have a difference of squares = 0
so
2(x+7) + (y-1) = 0 or 2(x+7) - (y-1) = 0
2x + 14 + y - 1 = 0 or 2x + 14 - y + 1 = 0
2x + y = -13 or 2x - y = -15
the "centre" would be the intersection of the two lines.
4x^2 + 56x - (y^2 - 2y ) = -195
4(x^2 + 14x + <b>49</b>) - (y^2 - 2y + <b>1</b>) = -195 + 4(49) - 1
4(x+7)^2 - (y-1)^2 = 0
ahh, we have a degenerated hyperbola
the hyperbola has degenerated into its asymtotes
notice we have a difference of squares = 0
so
2(x+7) + (y-1) = 0 or 2(x+7) - (y-1) = 0
2x + 14 + y - 1 = 0 or 2x + 14 - y + 1 = 0
2x + y = -13 or 2x - y = -15
the "centre" would be the intersection of the two lines.
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