Let A(0,1),B(1,1),and C(1,0) be the mid-point | vedclass

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(B) The mid-points of the sides of the triangle are $A(0,1)$,$B(1,1)$,and $C(1,0)$.Let the vertices of the triangle be $P, Q, R$. The mid-points are $

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$8$

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(B) The mid-points of the sides of the triangle are $A(0,1)$,$B(1,1)$,and $C(1,0)$.Let the vertices of the triangle be $P, Q, R$. The mid-points are $A = \frac{P+Q}{2}$,$B = \frac{Q+R}{2}$,$C = \frac{R+P}{2}$.Solving these,we get $P(0,0)$,$Q(0,2)$,and $R(2,0)$.The side lengths are $PQ = 2$,$QR = 2\sqrt{2}$,and $RP = 2$.The incentre $D$ is given by $\left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}\right)$.Here,$a=2\sqrt{2}$,$b=2$,$c=2$. Vertices are $(0,0), (0,2), (2,0)$.$D = \left(\frac{2\sqrt{2}(0) + 2(0) + 2(2)}{2\sqrt{2}+2+2}, \frac{2\sqrt{2}(0) + 2(2) + 2(0)}{2\sqrt{2}+2+2}\right) = \left(\frac{4}{4+2\sqrt{2}}, \frac{4}{4+2\sqrt{2}}\right) = \left(\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}\right)$.Rationalizing $D$,we get $\left(\frac{2(2-\sqrt{2})}{2}, \frac{2(2-\sqrt{2})}{2}\right) = (2-\sqrt{2}, 2-\sqrt{2})$.The parabola $y^2 = 4ax$ passes through $D(2-\sqrt{2}, 2-\sqrt{2})$.$(2-\sqrt{2})^2 = 4a(2-\sqrt{2})$ $\Rightarrow 4a = 2-\sqrt{2}$ $\Rightarrow a = \frac{2-\sqrt{2}}{4} = \frac{1}{2} - \frac{1}{4}\sqrt{2}$.The focus is $(a, 0) = (\frac{1}{2} - \frac{1}{4}\sqrt{2}, 0)$.Thus,$\alpha = \frac{1}{2}$ and $\beta = -\frac{1}{4}$.$\frac{\alpha}{\beta^2} = \frac{1/2}{(-1/4)^2} = \frac{1/2}{1/16} = 8$.

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

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