JEE Main 2026 — Jan 24, Shift 2 — Mathematics solution

JEE Main2026Jan 24, Shift 2Mathematics

Let the image of parabola $x^{2} = 4 y$ , in the line $x - y = 1$ be $(y + α)^{2} = b (x - c)$ , $a, b, c \in N$ . Then $a + b + c$ is equal to

Solution

A point on $x^{2} = 4 y$ is $P = (2 t, t^{2})$ . Reflect across $x - y - 1 = 0$ using the formula with $n = (1, - 1)$ , $∣ n ∣^{2} = 2$ :

$Q = P - \frac{2 ( 2 t - t ^{2} - 1 )}{2} (1, - 1) = P + (t - 1)^{2} (1, - 1)$

$Q = (2 t + (t - 1)^{2}, t^{2} - (t - 1)^{2}) = (t^{2} + 1, 2 t - 1)$ .

Let $(h, k) = (t^{2} + 1, 2 t - 1)$ . Then $t = \frac{k + 1}{2}$ and

$h = (\frac{k + 1}{2})^{2} + 1 \Rightarrow (k + 1)^{2} = 4 (h - 1)$ .

Locus: $(y + 1)^{2} = 4 (x - 1)$ . Matching $(y + α)^{2} = b (x - c)$ gives $α = 1$ , $b = 4$ , $c = 1$ .

$∴ a + b + c = 1 + 4 + 1 = 6$ .

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