Question

find the equation of the locus of all points equidistant from (3,5)and the x-axis​

Answer 1

Locus of the point is equal to x² - 6x - 10y +34 = 0

Let (x1,y1) be the point equidistant from the point (3,5) and the x-axis.

Distance from the point (3,5) = Distance from the x-axis

Distance of the point from (3,5) = √((x1-3)² + (y1-5)²)

Distance of the point from the x-axis = y1

=> √((x1-3)² + (y1-5)²) = y1²

=> x1² + 9 - 6x1 + y1² + 25 - 10y1 = y1²

=> x1² - 6x1 - 10y1 + 34 = 0

Locus of the point which is equidistant from the point (3,5) and the x - axis , x² - 6x - 10y +34 = 0