We must find all the points of tangency. 

I'll assume you haven't had calculus, it's a lots harder that way! lol. 

Let the common tangent line have equation y=mx+b, solve the system

 
and find the discriminant of the resulting quadratic to be 1-bm. Set it equal to
0, because there is only one solution at a tangent point.  Then solve the
system:
 
and find the discriminant of the resulting quadratic to be -b²+16m²+6.  Set it
equal to 0.  Then solve the system

You'll get


Substitute the positive values for b and m in y=mx+b, and solve simultaneously
with the parabola and get: 
  
and you can tell by symmetry that 

and with the ellipse and get:
  
and you can tell by symmetry that 


The quadrilateral CDBA is a trapezoid (or trapezium if you live in the UK).





The height is along the x-axis from -2 to 8 which is 10 units.



It's much easier If you've had calculus. Just equate the derivatives to find the
values of x where the slopes are the same.  You'll get the same answer.

Edwin