JEE Main 2026 — 24 January, Evening Shift — Question 6

Parametric point $P$ on $x^{2}=4 y$ is $P\left(2 t, t^{2}\right)$

∴ mirror image of P in $\mathrm{x}-\mathrm{y}=1$ is $\mathrm{Q} \equiv\left(2 \mathrm{t}-\frac{2 \cdot 1 \cdot\left(2 \mathrm{t}-\mathrm{t}^{2}-1\right)}{2}, \mathrm{t}^{2}+\frac{2 \cdot 2\cdot\left(2 \mathrm{t}-\mathrm{t}^{2}-1\right)}{2}\right)$

$\mathrm{Q} \equiv\left(\mathrm{t}^{2}+1,2 \mathrm{t}-1\right) \equiv(\mathrm{h}, \mathrm{k})$

∴ locus of Q is $\mathrm{x}=\frac{(\mathrm{y}+1)^{2}}{4}+1$

which is the required parabola. $\therefore(y+1)^{2}=4(x-1)$

$\therefore \mathrm{a}=1, \mathrm{~b}=4, \mathrm{c}=1$

$\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=6$