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