JEE Main 2023 — Straight Lines Question with Solution

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,

$I\equiv \left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$

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

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

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

$\Rightarrow a=\frac{1}{4+2\sqrt{2}}$

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

Hence, focus will be,$\left(\frac{1}{4+2\sqrt{2}},0\right)=\left(\frac{4-2\sqrt{2}}{8},0\right)$

Now comparing with $(\alpha +\beta \sqrt{2},0)$ we get,

$\alpha =\frac{4}{8}=\frac{1}{2}, \beta =-\frac{1}{4}$
Hence, $\frac{\alpha }{{\beta }^2}=\frac{\frac{1}{2}}{{\left(\frac{-1}{4}\right)}^2}=\frac{16}{2}=8$