Let the coordinates of point P be (x0, y0), then we can know x0 according to the coordinate characteristics of this point on the function image. +3y0？ = 4. First, according to the coordinates of point A, the coordinates of point B are (1,-1); Then, the formula1/2 | pa | is used to list the equation 1/2absinC. |PB|sin∠APB= 1/2|PM|？ | pn | sin ∠ MPN。 That is | pa |/| pm | = | pn |/| Pb |; Then according to the distance formula between two points, | x0+1| | 3-x0 | = | 3-x0 |/| x0-1|, namely (3-x0)? =|x0？ - 1|，x0 = 5/3。 It is easy to find the value of y0.

explain

Solution:

∫ point b and point a (-1, 1) are symmetrical about the origin o.

∴ The coordinates of point B are (1,-1).

If there is a point p that makes the areas of △PAB and △PMN equal.

The coordinates of point P are (x0, y0).

Then:

1/2|PA|？ |PB|sin∠APB

= 1/2|PM|？ |PN|sin∠MPN

∫sin∠APB = sin∠MPN

∴|PA|/|PM|=|PN|/|PB|

∴|x0+ 1|/|3-x0|=|3-x0|/|x0- 1|

Namely:

(3-x0)？ =|x0？ - 1|

Solution:

x0=5/3

∵ Point P is on the image of x2+3y2 = 4 (x ≠ 1).

∴x02+3y02=4

∴y0= √33/9

The existence of point p makes the areas of △PAB and △PMN equal.

At this time, the coordinate of point P is (5/3, √ 33/9).