Question

Find the equation of the locus of a point which moves so that its distances from the y axis is equal to the square of its distance from (h,0)

Answer 1

ok how do i begin this question?.. idk

Answer 2

@ganeshie8

Answer 3

@freckles

Answer 4

how do we define distance?

Answer 5

take a point (x,y)

what is its distance from the y axis?

what is its distance from (h,0) ??

equate them using the relationship criteria

Answer 6

Distance (x,y) to y axis = Distance (x,y) to (0,y) = x
Distance (x,y) to (h,0) = √ ( x - h)^2 + (y-0)^2
Directions say : Distance (x,y) to y axis = square of distance from (x,y) to (h,0)
x = [√ ( (x-h)^2 + (y-0)^2) ]^2
x = (x-h)^2 + y^2
(x-h)^2 -x + y^2 = 0
This is starting to look like an equation of a circle.

Complete the square
x^2 -2xh +h^2 -x + y^2 = 0
x^2 -2xh -x + y^2 = -h^2
x^2 -( 2h +1)x + y^2 = -h^2
( x - (2h+1)/2 ) ^2 + y^2 = -h^2 + ((2h+1)/2 ) ^2
( x - (2h+1)/2 ) ^2 + y^2 = h + 1/4

What do we know about this equation .
The locus of points is a circle with center ( (2h+1)/2, 0 ) and a radius √ (h + 1/4)
test this is true, plug in h = 6 for example , and graph it using a graphing utility or by hand
I would recommend Desmos calculator, as it lets you use a slider to check different values of h

https://www.desmos.com/calculator/ftm9ndy1vj

Verifying it graphically is like having someone double check your work for free.

Note that for h < -1/4 there is no meaningful output, can you see why?

Answer 7

http://www.wolframalpha.com/input/?i=sqrt%28x^2%29%3D%28sqrt%28%28x-%28-2%29%29^2%2B%28y%29^2%29%29^2

it works fine for h=-2