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JEE Advanced2006MathematicsApplication of DerivativesActual

Tangent is drawn at any point P of a curve which passes through (1,1) cutting X -axis and Y -axis at A and B , respectively. If A P: B P=3: 1 , then

Options

  1. Adifferential equation of the curve is 3 x d y d x +y=0
  2. Bdifferential equation of the curve is 3 x d y d x -y=0
  3. Ccurve is passing through ( 1 8 , 2 )
  4. Dnormal at (1,1) is x+3 y=4

Correct answer

A. differential equation of the curve is 3 x d y d x +y=0

Step-by-step solution

Since, A P: B P=3: 1 where, equation of tangent is y-y₁=f^ (x) (x-x₁ ) aligned & A (x₁- y₁ f^ (x₁ ) , 0 ) and B (0, y₁-x₁ f^ (x) ) & A P: B P=3: 1 d y d x = y -3 x & 3 x d y d x +y=0 or d y y =- 1 3 x d x, aligned On integrating both sides, we get array lrlr & x y^3 & =1 & At & x & = 1 8 and y=2 array

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