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NDA 1 2025 Maths, question 51

Topic: Analytical geometry

Q51

Under what condition will the lines m2x+ny−1=0m^{2}x + ny - 1 = 0 and n2x−my+2=0n^{2}x - my + 2 = 0 be perpendicular?

  1. (a) mn−1=0mn - 1 = 0Correct answer
  2. (b) mn+1=0mn + 1 = 0
  3. (c) m+n=0m + n = 0
  4. (d) m−n=0m - n = 0

Solution.

Answer: (a) mn−1=0mn - 1 = 0

  1. Two genuine lines are perpendicular when the dot product of their normal vectors is zero.
  2. Here this gives (m2,n)⋅(n2,−m)=m2n2−mn=mn(mn−1)=0\left( m^{2},n \right) \cdot \left( n^{2}, - m \right) = m^{2}n^{2} - mn = mn(mn - 1) = 0.
  3. Among the listed conditions, mn−1=0mn - 1 = 0 guarantees perpendicularity.
  4. It also guarantees that both lines are nondegenerate.