Lagrange - Minimising area under parabola, my answer is wrong

Let a>b>0. The area under $y=ax-bx^2$ between roots $x=0, x=\frac{a}{b}$ is $\frac{a^3}{6b^2}$.Find $a%$ and $b$, so that parabola passes through $(1,1)$ and the area is minimised.

My Attempt:

We have $y=ax-bx^2$ through point $(1,1) \implies (1)=a(1)-b(1)^2 \implies a-b=1 \implies a-b-1 = 0$

We want to minimise $\frac{a^3}{6b^2}$ subject to $a-b-1=0$, so let the lagrange be

L = $\frac{a^3}{6b^2}+\lambda(a-b-1)$

Find the first-order partial derivatives:

$\frac{\partial L}{\partial a}=\frac{a^2}{2b^2}+\lambda=0$

$\frac{\partial L}{\partial b}=-\frac{a^3}{3b^3}-\lambda=0$

$\frac{\partial L}{\partial \lambda} = a-b-1=0$

Then, using wolfram alpha I solved the system of equations giving:

$a=0, b=-1, \lambda = 0$

$a=1, b=0, \lambda = 0$

$a=\frac{1}{2}(1-\sqrt7), b = \frac{1}{2}(-1-\sqrt7), c = -\frac{1}{32}(\sqrt7 - 1)^2(1+\sqrt7)^2$

and $a=\frac{1}{2}(1+\sqrt7), b = \frac{1}{2}(\sqrt7 - 1), c = \frac{1}{32}(\sqrt7 - 1)^2(1+\sqrt7)^2$

From this the first 2 solutions do not satisfy $a>b>0$, so I am left with the final solution, however when I graphed this on geogebra it did not intercept the point $(1,1)$, so I must have made a mistake?

Any ideas what I did wrong? Many thanks