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COMEDK2017MathematicsApplication of Derivatives

Find the equations of the tangent and the normal to the curve y= x³ 4-x at (2,4)

Options

  1. A8 x+y-12=0 ; x+8 y+34=0
  2. B8 x-y-12=0 ; x-8 y-34=0
  3. C8 x+4 y+12=0, x-8 y+34=0
  4. D8 x-y-12=0, x+8 y-34=0

Correct answer

D. 8 x-y-12=0, x+8 y-34=0

Step-by-step solution

Given curve, y= x³ 4-x ...(i) is the given curve Differentiating Eq. (i) w.r.t. x , we get aligned d y d x =& (4-x) (3 x² )-x³(-1) (4-x)² = 12 x²-2 x³ (4-x)² ( d y d x )_ x=2 =& 12(2)²-2(2)³ (4-2)² = 32 4 =8 aligned The slope of tangent at (2,4) is 8 . The equation of tangent is gathered y-4=8(x-2) 8 x-y-12=0 slope = -8 -1 =8 gathered Now the slope of normal at P(2,4) is -1 8 . The equation of normal is aligned &y-4= -1 8 (x-2) & x+8 y-34=0 aligned

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