Analysis: (1) Let the moving point be m and the coordinates be (x, y), and find the straight line A? , horse? The slope of m, and their product, can get the trajectory equation of point m, discuss m according to the standard equation form of circle, ellipse and hyperbola, and determine the shape of curve; (2) According to (1), when m=- 1, the equation of C 1 is x? +y？ =a？ When m ∈ (- 1, 0)∩(0, +∞), the focus of C2 is f1(-a √1+m √, 0), and the necessary and sufficient conditions for F2 (a √) are ① xο +yο？ =a？
②﹙ 1/2﹚2a√﹙ 1+m﹚| y0 | = | m | a？ The coordinates of n points can be obtained, and the value of tanF 1NF2 can be obtained by using the product of quantities and the triangle area formula.
Solution: Solution: (i) Let the fixed point be m and the coordinates be (x, y).
When x ≠ a, kMAkMA can be obtained from the condition? =y/ ﹙x-a ﹚？ y/﹙ x+a ﹚=m，
Is that mx? -Really? = Horse? (x≠ a)，
A again? (-a，0)，A？ The coordinates of (a, 0) satisfy mx? -Really? = Horse? .
When m
When m=- 1, the equation of curve c is x? +y？ =a？ C is a circle whose center is at the origin;
When-1 < m < 0, the equation of curve c is x? /a？ +y/-Mom? (= 1, c is an ellipse whose focus is on the x axis;
When m > 0, the equation of curve c is x? /a？ +y/-Mom? (= 1, c is a hyperbola whose focus is on the x axis;
(2) According to (1), when m=- 1, the equation of C 1 is x? +y？ =a？ ,
When m ∈ (- 1, 0)∩(0, +∞), the focus of C2 is f1(-a √-1+m, 0) respectively.
F2 (a √ |1+m |, 0),
For a given m ∈ (- 1, 0)∩(0, +∞), there is a point N(xο, yο)(yο≠0) on C 1, so that △ f/. ,
The necessary and sufficient condition of is xο+yο=a? ①( 1/2)*2a√﹙ 1+m﹚| y0 | = | m | a？ ②
0 < | y0 |≤ a in ① and | y0 | = | m | a √ (1+m) in ②,
When 0 < | m | a/√1+m √≤ a, that is, √1-√ 5 √/2 ≤ m < 0, or 0 < m ≤
There is a point n that makes S=|m|a? ,
When | m | a/√1+m √ a, that is-1< m < √1-√ 5 √/2, or m > √.
When m ∈, there is a point n on C 1, which makes the area of △F 1NF2 S=|m|a? , and tanθ= 2；;
When m∈(0, ﹙ 1 √ 5 ﹚/2), there is a point n on C 1 so that the area of △F 1NF2 is | m | a? , and tan θ =-2;
When (-1, ﹙1-√ 5/2) ∩ (﹙1﹢√ 5 ﹚/2, +∞), the conditions are not met.