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Formula of parameter equation in senior high school mathematics
This kind of problem, if you are not familiar with polar coordinates, must be transformed into rectangular coordinates to solve, and the topic also requires equations of rectangular coordinates.

ρ= 1 is a circle with the origin (pole) as the center and the radius of 1, and the corresponding rectangular coordinate equation is x? +y? = 1;

The rectangular coordinates of n are x=√2cos(π/4)= 1, y=√2sin(π/4)= 1, n (1,1);

(I) let M(xm, ym), xm? +ym? = 1, g coordinate (x, y), according to the relationship between vector addition and coordinates:

x=xm+ 1,y = ym+ 1; xm=x- 1,ym = y- 1; Substitute into the above formula:

(x- 1)? +(y- 1)? = 1, this is the C2 equation of G, and it is also a circle with the center of N( 1, 1) and the radius of 1!

(2) In this parametric equation, t is the distance from the point with coordinates (x, y) on a straight line to p (2,0), and this distance is directional (positive and negative). According to the second formula, y is directly proportional to t, so the y-axis component and y-axis of t can be defined as positive direction (upward).

So |PA|=|t 1|, |PB|=|t2|

|PN|=√[(2- 1)? +(0- 1)? ]=√2 & gt; 1, ∴P is outside the circle, n, PA and PB are in the same direction, and t 1 is the same sign as t2.

Substitute the parameter equation into the trajectory equation of C2:

(2-t/2- 1)? +(t√3/2- 1)? = 1

( 1-t/2)? +(t√3/2- 1)? = 1

1-t+t? /4+3t? /4-t√3+ 1= 1

t? -( 1+√3)t+ 1=0

According to Vieta's theorem:

t 1+t2= 1+√3

t 1t2= 1

It can be seen that t 1 and t2 are both positive numbers.

1/|PA|+ 1/|PB|

= 1/t 1+ 1/t2

=(t 1+t2)/(t 1t2)

= 1+√3