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Question 1058625: What values of the constant a,b and c make the ellipse 4x^2+y^2+ax+by+c=0 lie tangent to the x axis at the origin and pass through the point -1,2?What is the eccentricity of the ellipse?
Answer by solve_for_x(190)   (Show Source):
You can put this solution on YOUR website! Since the ellipse is tangent to the x-axis at the origin, the point (0, 0) is on the curve of the ellipse.
This gives:
4(0)^2 + (0)^2 + a(0) + b(0) + c = 0 --> c = 0
Since c = 0, it can be dropped from the equation.
Rearranging the equation gives:
4x^2 + ax + y^2 + by = 0
Completing the square for both x and y gives:
The above is the equation of an ellipse that is centered on the point (-a/2, -b/2).
Since the ellipse is tangent to the x-axis at the origin, the x-coordinate of the center
of this ellipse is 0, which means that -a/2 = 0, or a = 0.
Substituting a = 0 into the above equation then leaves:
Then, the point (-1, 2) is also known to be on the curve. Substituting x = -1 and y = 2
into the equation and solving for b gives:
2b = -8
b = -4
Solution: a = 0, b = -4, c = 0
Graph of the ellipse:
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