Let an ellipse E: X2A2+Y2B2 = 1 (A > B > 0), the length of its major axis is twice that of its minor axis, and it is not in the focus. ...

Solution: (1) Let the coordinates of the focus be (c, 0).

Then the coordinates of the intersection of the straight line passing through the focus and perpendicular to the X axis and the ellipse are (c, y0).

Substitute x2a2+y2b2= 1, y0= b2a,

Because the chord length of the straight line passing through the focal point and perpendicular to the X axis is 23,

So 2×b2a=23,

From the meaning of the question, a=2b, substituted in the above formula, a=23, b=6,

So the elliptic equation is X2 12+Y26 = 1.

(2) suppose there is a point M(m, 0) (0 < m < 6) on the straight line of 2,

The straight line is not perpendicular to the x axis,

Let the equation of line L be y = k (x-6) (k ≠ 0).

Let P(x 1, y 1), Q(x2, y2),

(1+2k2) x2-46k2x+12k2-12 = 0 can be obtained from x21y = k (x-6).

Then x 1+x2=46k2 1+2k2, x 1? x2 = 12k 2- 12 1+2k 2。

∴MP=(x 1-m, y 1), MQ=(x2-m, y2), PQ=(x2-x 1, y2-y 1), where x2-x1.

∵ A parallelogram with MP and MQ as adjacent sides is a diamond.

∴(MP+MQ)⊥PQ？ (MP+MQ)？ PQ=0。

∴(x 1+x2-2m)(x2-x 1)+(y 1+y2)(y2-y 1)=0.

∴x 1+x2-2m+k(y 1+y2)=0.

∴46k2 1+2k2-2m+k2(46k2 1+2k2-26)=0.

Simplified to m = 6k21+2k2 = 61k2+2 (k ≠ 0),

Then m ∈ (0 0,62)

There is a point M(m, 0) and m∈(0, 62) on the straight line of 2.