Area of the quadrilateral formed by drawing tangents at the ends of latusrecta of x24+y21=1 is

Given ellipse is x24+y21=1 Here a=2,b=1e=a2−b2a2=4−14=32 Now, end point of latusrectum =Pae,b2a=P3,12 Equation of tangent is S1=0⇒3x+2y=4 Area of quadrilateral =4×△AOB=4×162|3×2|=163