Let A ( 0, 1 ) , B ( 1, 1 ) and C ( 1, 0 ) be the mid-points of the...

Given,

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

Now finding the vertices of the triangle by using midpoint formula and plotting the diagram we get,

Now finding the incentre of the above triangle using the formula,

$I ≡ \left(\dfrac{a{x}_{1} + b{x}_{2} + c{x}_{3}}{a + b + c}, \dfrac{a{y}_{1} + b{y}_{2} + c{y}_{3}}{a + b + c}\right)$

We get, Incentre $D = \left(\dfrac{4}{4 + 2\sqrt{2}}, \dfrac{4}{4 + 2\sqrt{2}}\right)$

Now given parabola ${y}^{2} = 4ax$ passes through the incentre $D$ we get,

${\left(\dfrac{4}{4 + 2\sqrt{2}}\right)}^{2} = 4a\left(\dfrac{4}{4 + 2\sqrt{2}}\right)$

$⇒ a = \dfrac{1}{4 + 2\sqrt{2}}$

Now we know that focus of parabola is given by, focus $( a, 0 )$

Hence, focus will be,\left(\dfrac{1}{4 + 2\sqrt{2}}, 0\right) = \left(\dfrac{4 - 2\sqrt{2}}{8}, 0\right)$</p><p><br />Now comparing with$( \alpha + \beta\sqrt{2}, 0 )$we get,</p><p>$\alpha = \dfrac{4}{8} = \dfrac{1}{2}, \beta = - \dfrac{1}{4}$<br />Hence,$\dfrac{\alpha}{{\beta}^{2}} = \dfrac{\dfrac{1}{2}}{{\left(\dfrac{- 1}{4}\right)}^{2}} = \dfrac{16}{2} = 8