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Manipal MET2017MathematicsApplication of Derivatives

The equation of the tangent to the curve (1+x^2 ) y=2-x , where it crosses the x -axis, is

Options

  1. Ax+5 y=2
  2. Bx-5 y=2
  3. C5 x-y=2
  4. D5 x+y-2=0

Correct answer

A. x+5 y=2

Step-by-step solution

The given equation of curve is (1+x^2 ) y=2-x , it meets x -axis at (2,0) or it can be rewritten as y= 2-x 1+x^2 On differentiating w.r.t. x , we get aligned d y d x & = (1+x^2 )(-1)-(2-x)(2 x) (1+x^2 )^2 & = -1-x^2-4 x+2 x^2 (1+x^2 )^2 ( d y d x )_ (2,0) & = -1-4-8+8 25 =- 1 5 aligned Required equation of tangent is (y-0)=- 1 5 (x-2) x+5 y=2

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