JEE Main 18 March 2021 Shift 2 Solved Paper

Given, equation of ellipse is
‌

x2

27

+y2=1
Now, equation of tangent to the ellipse at
P(3√3‌cos‌θ,sin‌θ) is given by
‌

3√3‌cos‌θ⋅x

27

+sin‌θ⋅y=1
⇒‌‌‌

x‌cos‌θ

3√3

+y‌sin‌θ=1 ...(i)
X-intercept of Eq. (i) is x=3√3‌sec‌θ=OA (say)
Y-intercept of Eq. (i) is y=cosecθ=OB (say)
∴ Sum of intercepts =3√3‌sec‌θ+cosecθ=f(θ) (say)
⇒f′(θ)=3√3‌sec‌θ‌tan‌θ−cosecθ⋅cot‌θ
Put f′(θ)=0
⇒‌‌tan3θ=‌

1

3√3

⇒θ=

π

6

∵f′(θ) changes sign from negative to positive when moving from left to right.
∴f(θ) will attain minima at θ=‌

π

6