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

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Parabola

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 :

Answer: (B)

For

y^{2} = 16 x

a = 4

and a point is

(4 t^{2}, 8 t)

; the end

(16, 16)

corresponds to

t = 2

. The other end of a focal chord has parameter

- \frac{1}{t} = - \frac{1}{2}

, i.e.

(1, - 4)

. Dividing the join of

(16, 16)

and

(1, - 4)

internally in the ratio

5 : 2

from the first end gives

P = (\frac{5 ( 1 ) + 2 ( 16 )}{7}, \frac{5 ( - 4 ) + 2 ( 16 )}{7}) = (\frac{37}{7}, \frac{12}{7})

with

α + β = 7

; dividing from the other end gives

(\frac{82}{7}, \frac{72}{7})

with

α + β = 22

. The minimum value of

α + β

7

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option