Calculus Integral Question 
Question: Given the point (a, b) in the first quadrant, find the downward-opening parabola that passed through the point (a, b) and the origin such that the area under the parabola is a minimum.

I've tried to solve it. But the answer can't be right, since it leads to a upward-opening parabola.
Here's what I've done so far.
Step One:
Let $$y = cx^2 + dx + e$$ be the equation for the parabola.
Because the parabola has a root at the origin, $$e = 0$$ The function for parabola can be written as $$y = cx^2 + dx$$
Step Two:
We are given that the parabola passes through the point (a, b) in the first quadrant. Plugging the numbers into the function above, $$b = c(a)^2 + d(a)$$ $$\frac{b-a^2c}{a}=d $$
Therefore, $$y=cx^2+\frac{b-a^2c}{a}x$$
Step Three:
Next, to find the area under the parabola, use the integral: $$\int_0^n(cx^2+\frac{b-a^2c}{a}x)dx$$where n is the second root of the parabola (first root is the origin).
For n, $$0 = cx^2+\frac{b-a^2c}{a}x$$ $$0 = cx+\frac{b-a^2c}{a}$$ $$x=-\frac{b-a^2c}{ac} = n$$
Now, substitute the expression for n, $$\int_0^{-\frac{b-a^2c}{ac}} (cx^2+\frac{b-a^2c}{a}x)dx$$ which equals $$\frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2$$ which equals $$\frac{b-a^2c}{6a^3c^2}$$
It can be concluded that the area under the parabola has the function $$A=\frac{b-a^2c}{6a^3c^2}$$
Step Four:
To find the minimum area -- as requested, take the derivative of A. $$A'=\frac{6a^3c^2(-a)^2-(b-a^2c)(12a^3c)}{(6a^3c^2)^2}$$ $$=\frac{a^2c-2b}{6a^3c^3}$$ Find its root: $$0=\frac{a^2c-2b}{6a^3c^3}$$ $$0=a^2c-2b$$ $$c=\frac{2b}{a^2}$$
Since (a, b) is in the first quadrant, c is a positive number.
Can you tell me where have I gone wrong?
Thanks!

Please note
$\displaystyle A = \frac{c}{3}\bigl(-\frac{b-a^2c}{ac}\bigr)^3+\frac{b-a^2c}{2a}\bigl(-\frac{b-a^2c}{ac}\bigr)^2 = \frac{(b - a^2 c)^3} {6 a^3 c^2}$
Taking derivatative wrt. $c$ and equating to zero,
$\displaystyle A' = 0 = \frac{2(b-a^2c)^3 + 3ca^2(b-a^2c)^2}{6a^3c^3}$
This gives us $\displaystyle c = \frac{b}{a^2}, c = - \frac{2b}{a^2}$
