If the ellipse with eccentricity E1 and the hyperbola with eccentricity E2 have the same focus, and the end points of the major axis, the end points of the minor axis and the distance from the focus to an asymptote of the hyperbola form an equal ratio sequence, then (E1 ^ 2-1) / (E2 ^ 2-1)=

If the ellipse with eccentricity E1 and the hyperbola with eccentricity E2 have the same focus, and the end points of the major axis, the end points of the minor axis and the distance from the focus to an asymptote of the hyperbola form an equal ratio sequence, then (E1 ^ 2-1) / (E2 ^ 2-1)=

Let the major half axis of the ellipse be a, then the half focal length is C = E1 * a, and the minor half axis B = a √ (1-e1 & # 178;);
Let the real axis of hyperbola be m, then its half focal length C = E2 * m, real half axis n = m √ (E2 & # 178; - 1), asymptote my ± NX = 0;
If the distance between the end point (± a, 0), the vertex (0, ± b) and the right point (± C, 0) and the asymptote is an arithmetic sequence, then:
(m*b)²/(m²+n²)=[|n*a|/√(m²+n²)]*[|n*c|/√(m²+n²)],
That is (m * b) ² = (n * a) * (n * c);
Because (M / N) &# 178; = 1 / (E2 & # 178; - 1), B & # 178; / (a * c) = A & # 178; (1-e1 & # 178;) / (A & # 178; * E1) = (1-e1 & # 178;) / E1
(e1²-1)/(e2²-1) = -e1；