The distance between a moving point and a straight line x = 8 is twice as long as the distance between it and point a (2.0)

The distance between a moving point and a straight line x = 8 is twice as long as the distance between it and point a (2.0)

Let the moving point be p (x, y)
Then | X-8 | = 2 √ [(X-2) ^ 2 + y ^ 2]
Square: x ^ 2-16x + 64 = 4 (x ^ 2-4x + 4 + y ^ 2)
3x^2+4y^2-48=0
This is an ellipse

The distance from moving point P to point a (8,0) is twice that of point B (2,0)

Let P (x, y)
(x-8)^2+(y-0)^2=4(x-2)^2+4(y-0)^2
arrangement:
4x^2-16x+16+4y^3-x^2+16x-64-y^2=0
3x^2+2y^2=48
x^2/16+y^2/24=1
ellipse

If the distance from the moving point P to the point F (1,1) and the straight line 3x + y-4 = 0 is equal, then the trajectory equation of point P is
If the distance from the moving point P to the point F (1,1) and the straight line 3x + y-4 = 0 is equal, then the trajectory equation of point P is solved in detail

Since the point F (1,1) is on the line 3x + y-4 = 0
The distance from the moving point P to the point F (1,1) and the straight line 3x + y-4 = 0 should be equal
So the trajectory of point P must be a straight line passing through point F and perpendicular to the line 3x + y-4 = 0
That is to say, (1,1) is a vertical equation with respect to 3x + y-4 = 0
Solution
The trajectory equation of point P is
X-3Y+2=0

If the distance from the moving point P to the point F (1,1) and the straight line 3x + y-4 = 0 is equal, then the trajectory equation of point P is ()
A. 3x+y-6=0B. x-3y+2=0C. x+3y-2=0D. 3x-y+2=0

If point F (1,1) is on the straight line 3x + y-4 = 0, then the trajectory of point P is a straight line passing through point F (1,1) and perpendicular to the known straight line. Because the slope of the straight line 3x + y-4 = 0 is - 3, the slope of the straight line is 13. From the oblique form of the point, we know that the trajectory equation of point P is Y-1 = 13 (x-1), that is, x-3y + 2 = 0, so we choose B