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 :

Mathematics · Parabola

JEE Main 2026 — 21 January, Evening Shift — Question 6

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

Option A:
$22$
Option B:
$7$
Correct
Option C:
$5$
Option D:
$16$

Answer: B

Step-by-step solution

∵ parameter of point A is $\mathrm{t}=2$

⇒ Parameter of point $B$ is $t=-\frac{1}{2}$

⇒ Coordinates of B is $(1,-4)$

Case 1: A $(16,16)$

$\alpha=\frac{5+32}{7}=\frac{37}{7}$

$\beta=\frac{-20+32}{7}=\frac{12}{7}$

$\Rightarrow \alpha+\beta=7$

Case 2:

$\alpha=\frac{2+80}{7}$ ,

$\beta=\frac{-8+80}{7}$

$\alpha+\beta=22$

So minimum value of $\alpha+\beta=7$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 21 January, Evening Shift
Subject: Mathematics
Chapter: Parabola
Topic: Chords connected with a Parabola

Let y^2=12 x be the parabola with its vertex at O . Let P be a point on the parabola and A be a point on the x -axis such that angle…

Answer: B

Step-by-step solution

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