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

Question

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

Accepted Answer

$$
\begin{aligned}
& \text { So, } \mathrm{D} \equiv\left(\frac{4}{2+2+2 \sqrt{2}}, \frac{4}{2+2+2 \sqrt{2}}\right) \\
& \equiv\left(\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}\right) \\
& =\left(\frac{2}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}, \frac{2}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}\right) \\
& \equiv(2-\sqrt{2}, 2-\sqrt{2})
\end{aligned}
$$

$$
\begin{aligned}
& \because y^2=4 a x \\
& (2-\sqrt{2})^2=4 a(2-\sqrt{2}) \\
& \Rightarrow 4 a=2-\sqrt{2} \Rightarrow a=\frac{2-\sqrt{2}}{4} \\
& \Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{4}=\alpha+\beta \sqrt{2} \\
& \Rightarrow \alpha=\frac{1}{2}, \beta=\frac{-1}{4} \\
& \text { So, } \frac{\alpha}{\beta^2}=\frac{\frac{1}{2}}{\frac{1}{16}}=8
\end{aligned}
$$

Solutions · 1