> with(plots):

This is a demonstration file.

Let's find the area of the following conic:

> conic:=x^2+2*y^2+x*y=4;

> implicitplot(conic,x=-3..3,y=-3..3);

> sols:=solve(conic,y);

"sols" represents the two explicit functions of x that make up the ellipse. In the next line I am giving the name "top" to the top half of the ellipse.

> top:=sols[1];bottom:=sols[2];

Here's a picture in case you didn't believe!

> plot([top,bottom],x=-3..3,y=-3..3,thickness=2);

Use implicit differentiation to determine where the ellipse has vertical tangents. These (their coordinates) will give us the limits for the integration.

> derivative:=implicitdiff(conic,y,x);

> solve(4*y+x=0,y);

> xcoords:=solve(subs(y=-1/4*x,conic));

Now find the straight line through the points on "y = top" with the above x-coords.

> subs(x=xcoords[1],top);subs(x=xcoords[2],top);

> line:=y+1/7*sqrt(14)=-1/4*(x-4/7*sqrt(14));

> line:=solve(line,y);

> P1:=plot([top,line,bottom],x=-3..3,y=-3..3,colour=[red,blue,green],thickness=2):

> P2:=implicitplot({x=4*sqrt(14)/7,x=-4*sqrt(14)/7},x=-3..3,y=-3..3,colour=magenta,thickness=2):

> display(P1,P2);

The plot above makes it look like I have the right sraight line. The area between these two curves is a "straightforward" Cal II problem (straightforward to formulate but perhaps difficult to evaluate):

> half_area:=Int(top-line,x=xcoords[2]..xcoords[1])=int(top-line,x=xcoords[2]..xcoords[1]);

Since this is only half the area of the original ellipse we simply multiply by two to get:

> area:=2*(half_area);

End of Demo