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!

The reqyired parabola should be of the type $y=-kx^2+hx, k>0$ if it passes through $(a,b)$ then $h=(b+ka^2)/a$. So the eq. of the parabola is $$y=-kx^2+\frac{(b+ka^2)}{a}x$$
This cots $x$ axis at $x=0$, $x=x_0=a+b/(ak)$, so the area made by the parabola on x-axis is
$$A(k)=\int_{0}^{x_0}\left(-kx^2+\frac{(b+ka^2)x}{a}\right) dx, k>0$$
$$A(k)=\frac{ab}{2}+\frac{b^3}{6a^3k^2}+\frac{b^2}{2ak}+\frac{a^3k}{6}$$
$$\implies A'(k)=\frac{a^3}{6}-\frac{b^3}{2a^3k^3}-\frac{b^2}{2ak^2}=0$$
Check that have three roots: $$k=-\frac{b}{a^2},-\frac{b}{a^2},\frac{2b}{a^2}$$
Third one being positive is useful here. Also check that $A''(k)>0$. Thus,
$$A_{min}=A(2b/a^2)=\frac{9ab}{8}$$
