Coordinate Geometry – Parabola – JEE Main 24 Jan 2026 Shift 2

Solution:

Let $P(2t, t^2)$ be a general point on the parabola $x^2 = 4y$.

Let $(h, k)$ be the image of $P$ in the line $x – y – 1 = 0$.

$$\frac{h – 2t}{1} = \frac{k – t^2}{-1} = -2 \frac{(2t – t^2 – 1)}{1^2 + (-1)^2}$$

$$\frac{h – 2t}{1} = \frac{k – t^2}{-1} = -(2t – t^2 – 1)$$

$h – 2t = t^2 – 2t + 1 \Rightarrow h = t^2 + 1$

$k – t^2 = -(t^2 – 2t + 1)(-1) \Rightarrow k – t^2 = 2t – t^2 – 1 \Rightarrow k = 2t – 1$

From $k = 2t – 1$, we get $t = \frac{k+1}{2}$.

Substitute $t$ in equation for $h$:
$$h = \left(\frac{k+1}{2}\right)^2 + 1$$

$$x = \frac{(y+1)^2}{4} + 1$$

$$4(x-1) = (y+1)^2$$

Comparing with $(y+a)^2 = b(x-c)$:
$a = 1, b = 4, c = 1$.

$$a + b + c = 1 + 4 + 1 = 6$$

Ans. (3)