Final answer:
The locus of points where AX = 3BX forms a circle centered at (⅓, 0) with a radius √(⅙), where a point X satisfies the distance condition from A = (0,0) and B = (1,0).
Step-by-step explanation:
Finding the Locus of Points X
To find the locus of points X such that AX = 3BX, where A = (0,0) and B = (1,0), we can use geometric methods instead of trigonometry. Let's consider a point X with coordinates (x,y). The distance AX is the distance from A to X, which can be represented by the formula √((x-0)^2 + (y-0)^2) due to the Pythagorean theorem. On the other hand, the distance BX is the distance from B to X, which is √((x-1)^2 + (y-0)^2).
According to the given condition, AX = 3BX, we can set up the following equation:
√(x^2 + y^2) = 3√((x-1)^2 + y^2)
Squaring both sides of the equation to eliminate the square roots gives:
x^2 + y^2 = 9((x-1)^2 + y^2)
Expanding and simplifying further:
x^2 + y^2 = 9x^2 - 18x + 9 + 9y^2
Bringing all terms to one side yields:
8x^2 - 18x + 8y^2 + 9 = 0
After rearranging, we find the equation of a conic section, which is a circle centered at (⅓, 0) with a radius √(⅙) for any point X satisfying the condition AX = 3BX.