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  • 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 y2 = 4ax passing through D is<img alt="(\alpha+\beta

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 y2 = 4ax passing through D is(\alpha+\beta \sqrt{3}, 0) , where \alphaand \beta are rational numbers, then\frac{\alpha}{\beta^2} is equal to


 

Option: 1

6


Option: 2

8


Option: 3

\frac{9}{2}


Option: 4

12


Answers (1)

best_answer

\begin{aligned} & \mathrm{a}=\mathrm{OP}=2 \quad \mathrm{~b}=\mathrm{OQ}=2 \quad \mathrm{c}=\mathrm{PQ}=2 \sqrt{2} \\ & (2,0) \quad(0,2) \quad(0,0) \\ & \mathrm{D}\left(\frac{4}{2+2+2 \sqrt{2}}, \frac{4}{2+2+2 \sqrt{2}}\right) \equiv \mathrm{D}\left(\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}\right) \\ & y^2=4 a x \Rightarrow\left(\frac{2}{2+\sqrt{2}}\right)^2=4 a \cdot\left(\frac{2}{2+\sqrt{2}}\right) \\ & \therefore 4 a=\frac{2}{2+\sqrt{2}} \therefore a=\frac{1}{2} \cdot \frac{2-\sqrt{2}}{4-2}=\frac{1}{4}(2-\sqrt{2}) \\ & \therefore \alpha=\frac{2}{4}=\frac{1}{2} \quad \beta=\frac{-1}{4} \\ & \therefore \frac{\alpha}{\beta^2}=8 \text { Ans } \\ & \end{aligned}

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Sanket Gandhi

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