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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2026 · 21 January · Evening Shift · Mathematics · Parabola

Let one end of a focal chord of the parabola y^ 2 =16 x be (16,16)...

Let one end of a focal chord of the parabola y2=16xy^{2}=16 x be (16,16)(16,16). If P(α,β)\mathrm{P}(\alpha, \beta) divides this focal chord internally in the ratio 5:25: 2, then the minimum value of α+β\alpha+\beta is equal to :

Options

  1. A1616
  2. B55
  3. C77
  4. D2222

Correct answer

C. 77

Step-by-step solution

Parabola y2=16xy^2 = 16x, a=4a = 4, focus at (4,0)(4, 0). Point (16,16)(16, 16) has parameter t1=2t_1 = 2.
Other end of focal chord: t2=−1/t1=−1/2t_2 = -1/t_1 = -1/2, giving (4⋅1/4,8⋅(−1/2))=(1,−4)(4 \cdot 1/4, 8 \cdot (-1/2)) = (1, -4).
P divides the chord in ratio 5:2 internally (two cases):
From (16,16)(16,16) to (1,−4)(1,-4): P=(377,127)P = \left(\frac{37}{7}, \frac{12}{7}\right), α+β=7\alpha + \beta = 7.
From (1,−4)(1,-4) to (16,16)(16,16): P=(827,727)P = \left(\frac{82}{7}, \frac{72}{7}\right), α+β=22\alpha + \beta = 22.
Minimum value of α+β=7\alpha + \beta = 7.
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