Given that the two points in the Cartesian coordinate plane are a (2,2), B (- 1, - 2), P is on the X axis, and PA = Pb, the coordinates of point P are obtained

Given that the two points in the Cartesian coordinate plane are a (2,2), B (- 1, - 2), P is on the X axis, and PA = Pb, the coordinates of point P are obtained

Let AB be a straight line y = KX + B, a (2,2), B (- 1, - 2) satisfy y = KX + B, then there are 2 = 2K + B, - 2 = - K + B, the solution is k = 4 / 3, B = - 2 / 3, that is, y = 4 / 3x-2 / 3, ∵ P on AB, ∵ P satisfies y = 4 / 3x-2 / 3, when y = 0, x = 1 / 2, that is, P (1 / 2,0)

Given two points a (- 3,0) B (3,0), if | PA | + | Pb | = 10, can we find the trajectory equation of P?
Writing process

In the plane rectangular coordinate system, the standard equation of the circle with O (a, b) as the center and R as the radius is (x-a) + (y-b) = R. the general equation of the circle: expand the standard equation of the circle, shift the term, merge... Let y = B, find out the two X values X1 and X2 at this time, and specify X1 & lt; X2, then: when x = - C / A & lt; X1 or x = - C / A & gt

Given two points a (0, - 3) and B (0,3), if | PA | - | Pb | = 4, then the trajectory equation of point P is

The solution consists of two known points a (0, - 3) and B (0,3) if | PA | - | Pb | = 4
And 4 ＜ / AB/
So the locus of P is a branch of hyperbola with a (0, - 3) and B (0,3) as focus,
And the branch is above the x-axis,
So 2A = 4, that is, a = 2
From 2C = 6, that is, C = 3
So B ^ 2 = C ^ 2-A ^ 2 = 5
So the trajectory equation of point P is
y^2/4-x^2/5=0(y＞0)

If the moving point P moves on the curve y = x ^ 2 + 1 and a (1,0), then the trajectory equation of the midpoint of PA is obtained

Let P coordinate be (m, n)
The coordinates (x, y) of the midpoint of PA are
x = (m + 1)/2
y = (n + 0)/2 = n/2
convert to
m = 2x - 1
n = 2y
P is on a known curve, so
n = m^2 + 1
therefore
2y = (2x -1)^2 + 1
Simplify to
y = 2x^2 + 2x + 1
It is also a parabola