JEE Main 2025 · 24 January · evening · Single correct · +4 / −1

The equation of the chord, of the ellipse x²/25+y²/16=1, whose mid-point is (3,1) is :

  1. (a) 5 x+16 y=31
  2. (b) 48 x+25 y=169
  3. (c) 4 x+122 y=134
  4. (d) 25 x+101 y=176
Answer and solution

Answer (b) 48 x+25 y=169

Solution

Maths sir · 3 steps

Midpoint given? Write T = S₁ at (3,1), and the chord equation comes out in one line.

  1. Use T = S1. For S: x²/a²+y²/b²-1=0, the chord with midpoint (x₁, y₁) is T = S₁. Here T = (x x₁)/a²+(y y₁)/b²-1 and S₁ = x₁²/a²+y₁²/b²-1, and the -1 cancels on both sides.

    3x/25+y/16 = 9/25+1/16

  2. Clear fractions. Multiply both sides by 400: 48x + 25y = 144 + 25 = 169. Its slope is -48/25.
    OptionPasses through (3,1)?Slope
    (a) 5x+16y=31yes, 15+16=31-5/16
    (b) 48x+25y=169yes, 144+25=169-48/25, matches
    (c) 4x+122y=134yes, 12+122=134-2/61
    (d) 25x+101y=176yes, 75+101=176-25/101
  3. So. Every option passes through (3,1), but only (b) has the right slope. The chord is 48x+25y=169, option (b).

The idea to keep. Chord with a known midpoint: T = S₁. Passing through the midpoint is not enough, because the slope -(b² x₁)/(a² y₁) is what fixes the chord.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 100 times in JEE Main since 2020. This question is from 2025.