Question 1058625
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:


{{{4(x^2 + ax) + (y^2 + by) = 0}}}


{{{4(x^2 + ax + (a/2)^2) + (y^2 + by + (b/2)^2) = 4(a/2)^2 + (b/2)^2}}}


{{{4(x + a/2)^2 + (y + b/2)^2 = a^2 + b^2/4}}}


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:


{{{4x^2 + (y + b/2)^2 = b^2/4}}}


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:


{{{4(-1)^2 + (2 + b/2)^2 = b^2/4}}}


{{{4 + 4 + 2b + b^2/4 = b^2/4}}}


{{{8 + 2b = 0}}}


2b = -8


b = -4


Solution: a = 0, b = -4, c = 0


Graph of the ellipse:


*[illustration Ellipse_1058625.png]