Question

Find the equation of tangent and normal line to the curve x² + 4x - 1 = y at the point (1,4).

Answer 1

slope of tangent = dy/dx = 2x + 4

slope of tangent at ( 1, 4 )  = 2 ( 1 ) + 4 = 6

Equation of Tangent whos slope is 6 and passing through (1, 4 ) will be

(y - 4)/( x -1 ) = 6

y - 4 = 6x - 6

6x - y - 2 = 0 is the Equation of Tangent ...

Slope of Normal at (1, 4 ) will be  = -1/6

Equation of Normal whos slope is -1/6 and passing through (1, 4 ) will be

(y-4)/(x-1) = -1/6

6 ( y - 4 ) =  - ( x - 1)

x + 6y - 5 =  0  is  Equation of Normal