AP EAPCET 23-Aug-2021 Shift 1 Paper

Given θ=

π

3

and x=a(θ−sin‌θ),y=a(1−cos‌θ)
Differentiate it, w.r.t. ' θ '

dx

dθ

=a(1−cos‌θ).

dy

dθ

=a(sin‌θ)
Then,

dy

dx

=

a‌sin‌θ

a(1−cos‌θ)

=

sin‌θ

1−cos‌θ

(

dy

dx

)θ=

π

3

=

sin‌ π 3

1−cos‌ π 3

=

√3 2

1− 1 2

=√3
∴ Slope of tangent =√3
Length of subtangent =

(y)θ= π 3

m

=

a(1−cos‌ π 3 )

√3

=

a( 1 2 )

√3

=

a

2√3

...(i)
Length of subnormal=(y)θ=

π

3

.m=a(1−cos‌

π

3

).√3 =a(1−

1

2

)√3=

a√3

2

...(ii)
Sum of Eqs. (i) and (ii),

a

2√3

+

a√3

2

=

√3a

6

+

a√3

2

=

√3a+3√3a

6

=

4√3a

6

=

2√3a

3

=

2a

√3

∴ Sum of length of subtangent and sub-normal is

2a

√3

.