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KCET2016MathematicsApplication of Derivatives

The equation of the normal to the curve ( y (1+x² )=2-x ) where the tangent crosses ( x )-axis is

Options

  1. A( 5 x-y-10=0 )
  2. B( x-5 y-10=0 )
  3. C( 5 x+y+10=0 )
  4. D( x+5 y+10=0 )

Correct answer

A. ( 5 x-y-10=0 )

Step-by-step solution

Given curve, y (1+x² )=2-x (1) At x-axis, y=0 so Eq. (1) becomes 0=2-x x=2 So, the point is (2,0) Now, differentiate Eq. (1) with respect to x, we get y^ (1+x² )+y(2 x)=-1 At point (2,0) , we have y^ (1+2² )+0(2 2)=-1 y^ (5)=-1 y^ = -1 5 This is the slope of the tangent at point (2,0) So, slope of normal is - 1 y^ =5 Therefore, equation of normal is, y-0=5(x-2) y=5 x-10 5 x-y-10=0

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