JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

Let one end of a focal chord of the parabola $y^{2} = 16 x$ be $(16, 16)$ . If $P (α, β)$ divides this focal chord internally in the ratio 5 : 2, then the minimum value of $α + β$ is equal to :

A
22
B
7
C
5
D
16

Solution

For $y^{2} = 16 x$ , $4 a = 16 \Rightarrow a = 4$ . Use parametric form $(a t^{2}, 2 a t) = (4 t^{2}, 8 t)$ .

Given one end is $(16, 16)$ ⇒ $t_{1} = 2$ . For a focal chord, $t_{1} t_{2} = - 1$ , so $t_{2} = - \frac{1}{2}$ , giving the other end as $B = (1, - 4)$ .

The internal-division ratio “5 : 2” can refer to either direction, giving two cases:

Case 1: $P$ divides $A (16, 16)$ to $B (1, - 4)$ in 5:2 (so $A P : P B = 5 : 2$ ). $α = \frac{5 ( 1 ) + 2 ( 16 )}{7} = \frac{37}{7}$ , $β = \frac{5 ( - 4 ) + 2 ( 16 )}{7} = \frac{12}{7}$ . $α + β = 7$ .

Case 2: $P$ divides $A$ to $B$ in 2:5. $α = \frac{2 ( 1 ) + 5 ( 16 )}{7} = \frac{82}{7}$ , $β = \frac{2 ( - 4 ) + 5 ( 16 )}{7} = \frac{72}{7}$ . $α + β = 22$ .

Minimum value: $7$ .

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