Ellipse

E:‌x2∕6+‌y2∕3=1, Tangent :y=m1x±√6m12+3
P:y2=12x, Tangent: y=m2x+‌

3

m2

For common tangent
m=m1=m2,±√6m12+3=‌

3

m2

⇒m=±1
⇒ equation of common tangents y=x+3 and y=−x−3 point of contact for parabola is (‌

a

m2

,‌

2a

m

)
⇒A1≡(3,6),‌‌A4(3−6)
Let A2(x1,y1)⇒ tangent to E is ‌

xx1

6

+‌

yy1

3

=1
A3 is mirror image of A2 in x-axis ⇒A3(−2,−1)Intersection point of T1=0 and T2=0 is (−3,0)
Area of quadrilateral A1A2A3A4=‌

1

2

(12+2)×5=35 square units