(1) Find the equation of ellipse C 1;
(2) Let the left focus of ellipse C 1 be F 1, the right focus be F2, the straight line L 1 passes through the point F 1 and is perpendicular to the long axis of ellipse 1, the moving straight line L2 is perpendicular to the point p, and the middle perpendicular of the straight line PF2 intersects the straight line L2 at the point m, thus finding out the point m.
(1) analysis: ∫ ellipse c1:x 2/a 2+y 2/b 2 =1(a > b > 0)，e = c/a =√2/2 = = & gt； a=√2c
The straight line y=x-2√2 is tangent to the circle x 2+y 2 = b 2.
∴ The radius of the circle b = | x-y-2 √ 2 |/√ 2 = 2 = = > a^2-b^2=c^2==>； c^2=b^2,a^2=8
The equation C 1 of the ellipse C 1 is x 2/8+y 2/4 =1.
(2) Analysis: ∵ ellipse c1:x 2/8+y 2/4 =1,left focus is f1(-2,0), and right focus is F2 (2,0).
∵ line L 1⊥X axis, ∴ line L 1: X =-2.
Let M(x, y)
∵ line L2⊥ line L 1 in ∴p(-2,y). p
∴ the midpoint of f2p is D(0, y/2), and the slope of f2p is-y/4 = = >; The slope of the vertical line of F2P is 4°/y y.
∴F2P The equation of the middle vertical line is y-y/2 = 4/y * x = = > x=y^2/8
∵M is the intersection point of the perpendicular line L2 of the line segment PF2.
The equation of locus C2 of point ∴M is y 2 = 8x.