Let $$A(0, 1)$$, $$B(1, 1)$$ and $$C(1, 0)$$ be the mid-points of the sides of a triangle with incentre at the point $$D$$. If the focus of the parabola $$y^2 = 4ax$$ passing through $$D$$ is $$\left(\alpha + \beta\sqrt{2}, 0\right)$$, where $$\alpha$$ and $$\beta$$ are rational numbers, then $$\frac{\alpha}{\beta^2}$$ is equal to

·  Find Vertices: Midpoints are $$A(0,1), B(1,1), C(1,0)$$. The vertices of the large triangle are $$V_1(0,0), V_2(0,2), V_3(2,0)$$.

• This is a right-angled isosceles triangle with legs of length 2.
• With $$a=2$$: $$D = (\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}) = (2-\sqrt{2}, 2-\sqrt{2})$$.
• $$(2-\sqrt{2})^2 = 4a(2-\sqrt{2}) \implies 4a = 2-\sqrt{2} \implies a = \frac{2-\sqrt{2}}{4}$$.
• $$\alpha = \frac{1}{2}$$, $$\beta = -\frac{1}{4}$$.
• $$\frac{1/2}{(-1/4)^2} = \frac{1/2}{1/16} = \mathbf{8}$$.

·  Find Incentre $$D$$: For a right-angled triangle $$(0,0), (a,0), (0,a)$$, the incentre is $$(\frac{a}{2+\sqrt{2}}, \frac{a}{2+\sqrt{2}})$$.

·  Find Parabola $$a$$: $$y^2 = 4ax$$ passes through $$D$$:

·  Identify Focus: Focus is $$(a, 0) = (\frac{1}{2} - \frac{1}{4}\sqrt{2}, 0)$$.

·  Calculate $$\frac{\alpha}{\beta^2}$$:

• $$\frac{1/2}{(-1/4)^2} = \frac{1/2}{1/16} = \mathbf{8}$$.