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KCET2010MathematicsApplication of Derivatives

P is the point of contact of the tangent form the origin to the curve y= _ e x . The length of the perpendicular drawn form the origin to the normal at P is

Options

  1. A1 2 e
  2. B1 e
  3. C2 e ²+1
  4. De ²+1

Correct answer

D. e ²+1

Step-by-step solution

Given, curve y = _ e x ...(i) Let the coordinate of point of contact P( , ) dy dx = 1 x Now, equation of tangent at ' P ' (y- )= 1 (x- ) Since, the tangent passing through the origin ie, (0,0) (0- )= 1 (0- ) =1 At 'P' from Eq. (i) aligned &= _ e 1 &= _ e aligned ( =1) _ e = _ e e = e So, point of contact is P(e, 1) . Now, slope of normal dy dx =- x ( d y d x )_ at (P) =-e Equation of normal at ' P ' gathered (y-1)=-e(x-e) y-1=-e x+e² e x+y- (e²+1 )=0 ...(ii) gathered The length of perpendicular drawn from the origi

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