1 For the ellipse the focus is given by (f,0) where f=√a²-b². For the hyperbola the focus is (f,0) where f=√(a²-2b²)+b²=√a²-b², as for the ellipse, so both conics are confocal.

2 If we write the equation for the ellipse y²/b²=1-x²/a² and for the hyperbola y²/b²=x²/(a²-2b²)-1, then we can equate the expressions in x: 1-x²/a²=x²/(a²-2b²)-1. Multiply through by a²(a²-2b²) we get: 2a²(a²-2b²)-x²(a²-2b²)=a²x², from which x²=a²(a²-2b²)/(a²-b²). So x²/a²=(a²-2b²)/(a²-b²) and y²/b²=1-(a²-2b²)/(a²-b²)=b²/(a²-b²) and y=b²/√(a²-b²) [which can also be written b²/C if C²=a²-b²].

There are four points of intersection:

(a√((a²-2b²)/(a²-b²)), b²/√(a²-b²));

(a√((a²-2b²)/(a²-b²)), -b²/√(a²-b²));

(-a√((a²-2b²)/(a²-b²)), b²/√(a²-b²));

(-a√((a²-2b²)/(a²-b²)), -b²/√(a²-b²)).

3 The first quadrant intercept is at x=a√((a²-2b²)/(a²-b²))=a√(a²-2b²)/C, y=b²/C.

b²=a²(1-e²) for the ellipse, so e=√(a²-b²)/a, and E=√((a²-2b²+b²)/(a²-2b²))=√((a²-b²)/(a²-2b²)) for the hyperbola. Therefore 1/E=√((a²-2b²)/(a²-b²)) and 1/e=a/√(a²-b²).

So 1/eE=a√(a²-2b²)/C²=x/C, therefore x=C/eE. The first quadrant intercept is therefore at x=C/eE and y=b²/C, QED.