The intersection p of GF 1 and l is on an ellipse,
∴2a=|pf 1|+|pf2|=|gf 1|=4.
∴b2=a2-c2=3.
So the elliptic equation is X24+Y23 = 1.
(II) It is known from the conditions that the slopes of the straight lines PM and PN exist and are not 0,
Easy to get point p (- 1, 3 2), let the equation of straight line PM be y = k (x+1)+3 2,
Y is eliminated simultaneously by elliptic equation and linear PM equation,
Arrange (4k2+3) x2+4k (2k+3) x+4k2+12k-3 = 0,
∵P On the ellipse, the two roots of the equation of∴ are 1 and x 1.
∴ 1? x 1=？ 4k2+ 12k？ 3 4k2+3，x 1=？ 4k2+ 12k？ 3 4k2+3，
The inclination angles of the lines PM and PN are complementary,
The slopes of the straight lines PM and PN are opposite to each other.
∴x2=？ 4k2？ 12k？ 3 4k2+3。
So x 1? x2=？ 24k 4k2+3，x 1+x2=6？ 8k2 4k2+3。
And y1= k (x1+1)+32, y2 =? k(x2+ 1)+3 2，
∴y 1-y2=k(x 1+x2+2)= 12k 4k 2+3。
The slope of a straight line MN kmn = y 1? y2 x 1？ x2 =？ 12 (fixed value)