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).