Ellipse

Point of intersection of tangents at P and Q is R (2 secθ, 0)
Area of ∆PQR =

1

2

‌⋅‌2√3sin‌θ‌⋅‌(2‌sec‌θ‌−‌2‌cos‌θ)
⇒ ∆ = 2√3‌⋅‌

sin3θ

cos‌θ

; where cos‌θ‌∈[

1

4

,

1

2

]
Now
As θ increases, ∆ increases ⇒ when cos‌θ decreases, ∆ increases
∴ ‌‌‌∆min occurs at cos‌θ=

1

2

, Therefore
∆max occurs at cos‌θ=

1

4

, Therefore ∆1=2√3⋅

(1− 1 16 )3∕2

1∕4

= 8√3⋅

15⋅√15

4⋅4⋅4

=

2√3⋅15⋅√3√5

16

⇒ ‌‌∆1=

45

8

√5
Now

8

√5

‌∆1−8∆2=45−36=9