The required point is point (2, 3)
Let the equation of the tangent line be, y = mx + c; where y = 3, m = 5 and x = 2
Thus, 3 = 5(2) + c = 10 + c
c = 3 - 10 = -7
Therefore, tangent equation is y = 5x - 7
Equation of the normal is the equation of the line perpendicular to the equation of the tangent at point (2, 3).
Thus, the slope of the normal equation is -1/5
Normal equation is y - 3 = -1/5(x - 2)
5(y - 3) = -(x - 2)
5y - 15 = -x + 2
x + 5y = 2 + 15
x + 5y = 17 is the normal equation.