Coordinate Geometry – Parabola – JEE Main 21 Jan 2026 Shift 2

Solution:

Parabola $y^2 = 16x \Rightarrow 4a = 16 \Rightarrow a = 4$.

Let $A(at_1^2, 2at_1) = (16, 16)$.
$$2(4)t_1 = 16 \Rightarrow 8t_1 = 16 \Rightarrow t_1 = 2$$

For a focal chord connecting points $t_1$ and $t_2$, we have $t_1t_2 = -1$.
$$2 \cdot t_2 = -1 \Rightarrow t_2 = -\frac{1}{2}$$

Coordinates of the other end $B(at_2^2, 2at_2)$:
$$B\left(4\left(-\frac{1}{2}\right)^2, 2(4)\left(-\frac{1}{2}\right)\right) = B(1, -4)$$

We have two cases for the division of the chord AB by point P in ratio 5:2.

**Case 1:** P divides segment AB in ratio 5:2 ($AP:PB = 5:2$).

$$A(16, 16), \quad B(1, -4)$$
$$\alpha = \frac{5(1) + 2(16)}{5+2} = \frac{5+32}{7} = \frac{37}{7}$$
$$\beta = \frac{5(-4) + 2(16)}{5+2} = \frac{-20+32}{7} = \frac{12}{7}$$
$$\alpha + \beta = \frac{37+12}{7} = \frac{49}{7} = 7$$

**Case 2:** P divides segment BA in ratio 5:2 ($BP:PA = 5:2$).

$$B(1, -4), \quad A(16, 16)$$
$$\alpha = \frac{5(16) + 2(1)}{5+2} = \frac{80+2}{7} = \frac{82}{7}$$
$$\beta = \frac{5(16) + 2(-4)}{5+2} = \frac{80-8}{7} = \frac{72}{7}$$
$$\alpha + \beta = \frac{82+72}{7} = \frac{154}{7} = 22$$

The minimum value of $\alpha + \beta$ is $\min(7, 22) = 7$.

Ans. (2)