It is given that
y = x4 – 6x3 + 13x2 – 10x + 5
By substituting the value of x = 1
y = (1)4 – 6(1)3 + 13(1)2 – 10(1) + 5
On further calculation
y = 1 – 6 + 13 – 10 + 5 = 3
Point of contact is (1, 3)
Consider y = x4 – 6x3 + 13x2 – 10x + 5 as the equation of curve
By differentiating w.r.t. x
dy/dx = 4x3 – 18x2 + 26x – 10
So we get
(dy/dx)(1, 3) = 4(1)3 – 18(1)2 + 26(1) – 10 = 2
Here the required equation of tangent is
y – y1 = m(x – x1)
By substituting the values
y – 3 = 2x – 2
On further calculation
2x – y – 2 + 3 = 0
So we get
2x – y + 1 = 0
Here the required equation of normal i
y – y1 = -1/m(x – x1)
By substituting the values
y – 3 = -1/2(x – 1)
On further calculation
2y – 6 = – x + 1
So we get
2y – 6 + x – 1 = 0
x + 2y – 7 = 0