find the equation of the path of a point that moves so that its distance from (5, 0) is five~fourths of its distance from the line x=16\3
let the point be P(x,y)
If A is (5,0) and B = (16/3,y), then
AP = (5/4)BP
4 AP = 5 BP
4√( (x-5)^2 + (y-0)^2) = 5 (16/3 - x)
4√( (x-5)^2 + y^2 = 5(16/3-x)
square both sides
16( (x-5)^2 + y^2) = 25(16/3 - x)^2
16(x-5)^2 + 16y^2 = 25(16/3 - x)^2
looks hyperbolic to me.
check my algebra and arithmetic
To find the equation of the path of a point that moves such that its distance from (5, 0) is five-fourths of its distance from the line x = 16/3, we can follow these steps:
1. Let's assume the coordinates of the moving point are (x, y).
2. The distance between the point (x, y) and (5, 0) can be calculated using the distance formula:
distance = sqrt((x - 5)^2 + y^2)
3. The distance between the point (x, y) and the line x = 16/3 can be calculated by taking the absolute difference between x and 16/3.
distance = abs(x - 16/3)
4. According to the given condition, the distance of the point from (5, 0) is five-fourths of its distance from the line x = 16/3. So, we can set up a proportion:
sqrt((x - 5)^2 + y^2) = (5/4) * abs(x - 16/3)
5. Square both sides of the equation to eliminate the square root:
(x - 5)^2 + y^2 = (25/16) * (x - 16/3)^2
6. Simplify and expand both sides of the equation:
x^2 - 10x + 25 + y^2 = (25/16) * (x^2 - (32/3)x + 256/9)
16x^2 - 160x + 400 + 16y^2 = 25x^2 - (25/3)x^2 - 800x + 12800/9
Simplify the equation further:
16x^2 - 25x^2 + 160x + 800x - 25x^2 - 12800/9 + 160y^2 - 400 = 0
-34x^2 + 960x + 160y^2 - 12800/9 - 400 = 0
-34x^2 + 960x + 160y^2 - 14400/9 = 0
-34x^2 + 960x + 160y^2 - 1600 = 0
7. Rearrange the terms to get the equation in the standard form:
-34x^2 + 960x + 160y^2 = 1600
Therefore, the equation of the path of the point is -34x^2 + 960x + 160y^2 = 1600.