AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Given, x=35‌sec‌θ...(i)
y=35‌tan‌θ..(ii)
∴ Point P=(35‌sec‌θ⋅35‌tan‌θ)
Differentiate Eq. (i) W.I.to ' θ '
‌

dx

dθ

=35(sec‌θ⋅tan‌θ) ....(iii)
Differentiate Eq. (ii) W.r.to ' θ '
‌

dy

dθ

=35(sec2θ) ...(iv)
⇒ ‌‌‌

‌ Eq. (iv) ‌

‌ Eq. (iii) ‌

⇒‌

dy

dx

=‌

35(sec2θ)

35(sec‌θ⋅tan‌θ)

‌

dy

dx

=‌

sec‌θ

tan‌θ

=‌

1

sin‌θ

∴‌‌m=‌

1

sin‌θ

Equation of tangent is
y−35‌tan‌θ=m(x−35‌sec‌θ)
y−35‌

sin‌θ

cos‌θ

=‌

1

sin‌θ

(x−35⋅‌

1

cos‌θ

)
y‌sin‌θ−35‌

sin2θ

cos‌θ

=x−‌

35

cos‌θ

y‌sin‌θ=x+‌

35sin2θ

cos‌θ

−‌

35

cos‌θ

y‌sin‌θ=x+‌

35(sin2θ−1)

cos‌θ

y‌sin‌θ=x+‌

35(−cos2θ)

cos‌θ

y‌sin‌θ=x−35‌cos‌θ