If normals at P, Q, R to the parabola $y^{2} = 12 x$ are concurrent at $(7, 1)$ and the centre of circle
through P, Q, R is at $(α, β)$ then $[α] + 4 β$ =_________ (where [.]= g.i.f)

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answer is 7.

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Detailed Solution

If (h, k) is point of concurrence of the three normal at P, Q, R on $y^{2} = 4 a x$ then circumcenter of $Δ P Q R$ is $(\frac{h + 2 a}{2}, \frac{k}{4})$

$α = \frac{13}{2}, β = \frac{1}{4}$

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