JEE Main 2023 — Mathematics Coordinate Geometry

Given,

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

$$I\equiv (\frac{a{x}_{1}+b{x}_{2}+c{x}_{3}}{a+b+c},\frac{a{y}_{1}+b{y}_{2}+c{y}_{3}}{a+b+c})$$I≡(a+b+cax1​+bx2​+cx3​​,a+b+cay1​+by2​+cy3​​)

We get, Incentre $D=(\frac{4}{4+2\sqrt{2}},\frac{4}{4+2\sqrt{2}})$D=(4+22​4​,4+22​4​)

Now given parabola ${y}^{2}=4ax$y2=4ax passes through the incentre $D$D we get,

$${(\frac{4}{4+2\sqrt{2}})}^{2}=4a(\frac{4}{4+2\sqrt{2}})$$(4+22​4​)2=4a(4+22​4​)

$$\Rightarrow a=\frac{1}{4+2\sqrt{2}}$$⇒a=4+22​1​

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

Hence, focus will be,$(\frac{1}{4+2\sqrt{2}},0)=(\frac{4-2\sqrt{2}}{8},0)$(4+22​1​,0)=(84−22​​,0)

Now comparing with $(\alpha +\beta \sqrt{2},0)$(α+β2​,0) we get,

$$\alpha =\frac{4}{8}=\frac{1}{2},\beta =-\frac{1}{4}$$α=84​=21​,β=−41​

Hence, $\frac{\alpha }{{\beta }^{2}}=\frac{\frac{1}{2}}{{(\frac{-1}{4})}^{2}}=\frac{16}{2}=8$β2α​=(4−1​)221​​=216​=8