Point Q moves on the curve X ^ 2 + y ^ 2 = 1, and the symmetric point of point Q about point a (1, - 1) is p. find the trajectory equation of P

Point Q moves on the curve X ^ 2 + y ^ 2 = 1, and the symmetric point of point Q about point a (1, - 1) is p. find the trajectory equation of P

Let Q (x0, Y0), P (x1, Y1)
Since Q and P are symmetric with respect to a (1, - 1)
So a is the midpoint of the QP line, so
1=(x0+x1)/2
-1=(y0+y1)/2
Then x0 = 2-x1, Y0 = - 2-y1
Because Q moves on the curve X ^ 2 + y ^ 2 = 1
So x0 ^ 2 + Y0 ^ 2 = 1
That is, (2-x1) ^ 2 + (- 2-y1) ^ 2 = 1
Then the trajectory equation of point P: (X-2) ^ 2 + (y + 2) ^ 2 = 1

Given that point P moves on the curve (Y-2) ^ 2 = 16 (2-x), and point Q and point P are symmetrical about point (1,1), then the trajectory equation of point q is

(y² =16x)
From the symmetry: (x + x) / 2 = 1 (y + y) / 2 = 1
So: x = 2-x, y = 2-y
Then, the above formula is substituted into the trajectory of point P
The final answer

Point P moves on the curve C: y = x ^ 2-1, and point a (2,0) extends PA to Q
Point P moves on the curve C: y = x ^ 2-1, fixed point a (2,0), extends PA to Q, so that | AQ | = 2 | AP |, then the trajectory equation of point q is

Let P (x1, Y1), Q point coordinates: (x, y), a (2,0)
PA extended to Q, | AQ | = 2 | AP|
The fixed ratio of PQ of a divided directed line segment is: PA / AQ = 1 / 2
∴(2-x1)/(x-2)=1/2
(0-y1)/(y-0)=1/2
x1=(x+6)/2,y1=-y/2
∵ P (x1, Y1) is on the curve C: y = x & # 178; - 1,
∴-y/2=[(x+6)/2]²-1
The trajectory equation of Q point is as follows
y=-x²/2 -6x -16

When a moving point P moves on the circle x 2 + y 2 = 1, the trajectory equation of the midpoint m connecting it with the fixed point a (3,0) is obtained

Take any point B (m, n) on the circle x2 + y2 = 1, and let the midpoint m (x, y) of line AB have x = 3 + M2y = 0 + N2, that is, M = 2x-3n = 2Y. According to M2 + N2 = 1, we can get (x-32) 2 + y2 = 14, that is, the trajectory equation of midpoint m is (x-32) 2 + y2 = 14