Let A(0,1),B(1,1) and C(1,0) be the mid-points of the sides of a triangle with incentre at…

Question

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 $(\alpha +\beta \sqrt{2},0)$, where $\alpha $ and $\beta $ are rational numbers, then $\frac{\alpha }{{\beta }^{2}}$ is equal to :

Accepted Answer

$\begin{gathered} D\equiv (\frac{4}{2+2+2\sqrt{2}},\frac{4}{2+2+2\sqrt{2}}) \ \equiv (\frac{2}{2+\sqrt{2}},\frac{2}{2+\sqrt{2}}) \ =(\frac{2}{2+\sqrt{2}}	imes \frac{2-\sqrt{2}}{2-\sqrt{2}},\frac{2}{2+\sqrt{2}}	imes \frac{2-\sqrt{2}}{2-\sqrt{2}}) \ \equiv (2-\sqrt{2},2-\sqrt{2}) \end{gathered}$

$\begin{gathered} \because {y}^{2}=4ax \ (2-\sqrt{2}{)}^{2}=4a(2-\sqrt{2}) \ \Rightarrow 4a=2-\sqrt{2}\Rightarrow a=\frac{2-\sqrt{2}}{4} \ \Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{4}=\alpha +\beta \sqrt{2} \ \Rightarrow \alpha =\frac{1}{2},\beta =\frac{-1}{4} \ \frac{\alpha }{{\beta }^{2}}=\frac{\frac{1}{2}}{\frac{1}{16}}=8 \end{gathered}$

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