Question

Find the equation of tangent and normal at x = -1 for the curve y = 2x 2 +3x -1

Answer 1

y=2×2+3x-1
y=2×2+3×-1-1
y=4-3-1
y=1-1
y=0

Answer 2

equation of curve is y = 2x² + 3x - 1
at x= -1,y = 2(-1)² + 3(-1) -1
           ⇒y = 2 - 3 - 1
          ⇒y= 2 - 4
           ⇒y= -2
(1)Now,slope of tangent i.e. dy/dx =4x + 3 = 4(-1) + 3 = -4 + 3 = -1 
             slope of tangent = -1
∴ Equation of tangent having slope -1 and passing through point (-1,-2) is            y - (-2) = -1[ x - (-1)]
             ⇒y + 2 = -1[x+1]
             ⇒y + 2 = -x -1
             ⇒y + x +2 - 1 = 0
              ⇒x + y + 3 = 0       
(2). slope of normal = -dx/dy = 1...................(∵dy/dx = -1)
  ⇒slope of normal = 1
∴Equation of normal having slope 1 and passing through point (-1,-2)
 is      y - (-2) = 1[x - (-1)]
      ⇒y + 2 = x + 1
     ⇒y - x + 1 = 0      Ans