KEAM 2010 Math Paper

Given that
x=a(1+cos‌θ)...........(i)
and y=a(θ+sin‌θ).....(ii)
Differentiating Eqs. (i) and (ii) w.r.t. θ, we get
‌

dx

dθ

=−a‌sin‌θ....(iii)
and ‌

dy

dθ

=a+a‌cos‌θ.....(iv)
Dividing Eq. (iii) by Eq. (iv), we get
‌

dy

dθ

/

dx

dθ

=‌

a(1+cos‌θ)

−a‌sin‌θ

⇒‌

dy

dx

=‌

1+cos‌θ

−sin‌θ

=−‌

1+2cos2 θ 2 −1

2‌sin‌ θ 2 ‌cos‌ θ 2

=−‌

cos‌ θ 2

sin‌ θ 2

=−cot‌

θ

2

∴‌

d2y

dx2

=cosec2‌

θ

2

⋅‌

1

2

‌

dθ

dx

=−‌

1

2

‌

1

a

‌

cosec2 θ 2

sin‌θ

∴(‌

d2y

dx2

)θ=‌

π

2

=−‌

1

a