Question 1149909
<pre>{{{drawing(400,256,-10,15,-8,8,
green(line(-12,-sqrt(2),12,5sqrt(2)),line(-12,sqrt(2),12,-5sqrt(2))),

graph(400,256,-10,15,-8,8,2sqrt(x)),
graph(400,256,-10,15,-8,8,-2sqrt(x)),

red(line(-2,2.121319634,-2,-2.121319634), line(8,5.656857377,8,-5.656857377),

line(-2,2.121319634,8,5.656857377),line(-2,-2.121319634,8,-5.656857377)), 
locate(7.8,6.6,A),locate(7.8,-5.7,B),

locate(-2,3.2,C), locate(-2,-2.3,D),

arc(0,0,8,-2sqrt(6)) )}}}

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

{{{system(y=mx+b,y^2=4x)}}} 
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:
{{{system(y=mx+b,3x^2+8y^2=48)}}} 
and find the discriminant of the resulting quadratic to be -b²+16m²+6.  Set it
equal to 0.  Then solve the system
{{{system(1-bm=0,-b^2+16m^2+6=0)}}}
You'll get
{{{matrix(1,7,    b="" +- 2sqrt(2),"","",and,"","",m="" +- sqrt(2)/4)}}}

Substitute the positive values for b and m in y=mx+b, and solve simultaneously
with the parabola and get: 
{{{A=(matrix(1,3,8,",",4sqrt(2)))}}}  
and you can tell by symmetry that 
{{{B=(matrix(1,3,8,",",-4sqrt(2)))}}}
and with the ellipse and get:
{{{A=(matrix(1,3,-2,",",3sqrt(2)/2))}}}  
and you can tell by symmetry that 
{{{B=(matrix(1,3,-2,",",-3sqrt(2)/2))}}}

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

{{{Area=expr(1/2)(b[1]+b[2])h}}}

{{{b[1]=6sqrt(2)/2=3sqrt(2)}}}
{{{b[2]=8sqrt(2)}}}
The height is along the x-axis from -2 to 8 which is 10 units.

{{{Area=expr(1/2)(3sqrt(2)+8sqrt(2))*10=55sqrt(2)}}}

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</pre>