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MHT CET202525 Apr 2025Evening ShiftMathematicsApplication of DerivativesActual

The length of the perpendicular drawn from the origin on the normal to the curve x^2+2 x y-3 y^2=0 at the point (2,2) is

Options

  1. A2 units
  2. B3 2 units
  3. C2 2 units
  4. D1 2 units

Correct answer

C. 2 2 units

Step-by-step solution

The perpendicular from the origin to the normal of the curve x^2 + 2xy - 3y^2 = 0 at (2,2) is determined. Implicit differentiation gives the slope of the tangent at any point: dy dx = x + y 3y - x At (2,2) , the slope of the tangent is m_t = 2 + 2 3(2) - 2 = 1 , so the slope of the normal is m_n = -1 . The equation of the normal is then y - 2 = -1(x - 2) or x + y - 4 = 0 . The perpendicular distance from the origin to this line is |1(0) + 1(0) - 4| 1^2 + 1^2 = 4 2 = 2 2 The length is 2 2 units, corresponding to opt

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