JEE Main 2023 — Parabola Question with Solution

Given,

Equation of parabola is $y^2=6x$,

Also the general equation of a parabola is $y^2=4ax$

On comparing, we get

$\Rightarrow 4\mathrm{a}=6\Rightarrow \mathrm{a}=\frac{3}{2}$

Tangent on parabola $y^2=4ax$ is given by $y=mx+\frac{a}{m}$,

So, $y=mx+\frac{3}{2 m}; (m\ne 0)$

Now using the midpoint formula we get,

$\mathrm{h}=\frac{6 \mathrm{m}-3}{4 {\mathrm{m}}^2}, \mathrm{k}=\frac{6 \mathrm{m}+3}{4 \mathrm{m}}$ 

Now, eliminating $\mathrm{m}$ and we get $3\mathrm{h}=2\left(-2{\mathrm{k}}^2+9\mathrm{k}-9\right)$

$\Rightarrow 3h=-4k^2+18k-18$

$\Rightarrow 4k^2-18k+3h+18=0$

$\therefore$ The locus of its circumcentre is

$4y^2-18y+3x+18=0$