Application of Derivatives

Given equation of curve is
y‌=x2+x−1‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
⇒‌‌‌

dy

dx

‌=2x+1
∴‌

dy

dx

‌

‌

|x=1‌=2×1+1=3=m
i.e. slope of tangent =m=3. and point is ( 1,1 ) (by putting x=1 in Eq. (i))
Now, length of tangent
‌=a=|‌

y√1+m2

m

|
‌=|‌

1√1+32

3

|=‌

√10

3

length of subtangent
=b=|‌

y

m

|=|‌

1

3

|=‌

1

3

length of normal
=c=|y√1+m2|=|1×√10|=√10
length of subnormal
=d=|y⋅m|=|3|=3