equation of a locus

Question

Find the equation of a locus of point whose distance from the point (2,-2) is equal to its distance from the line      x-y=0

 

 

Answer 1

First of all, note that it will be a parabola, and that (2,-2) is it's focus and x - y = 0 is it's directrix.

Let's work with the squares of the distances instead since then there won't be any radicals.

Given a point (x,y) on this parabola, we have the squared distance from the focus as

D²_focus = (x - 2)² + (y + 2)².

To get the squared distance from the directrix, D²_directricx , we we can use vectors.

Pick a vector perpendicular to the directrix.

Recall that vector < a , b > is perpendicular to line ax + by = c

For example, v = < 1 , -1 > is perpendicular to our directrix x - y = 0.

Let u = < x , y >.

Then the desired distance is D_directricx = u · v / ||v||.

Thus D²_directricx = (<x , y> · <1 , -1> / || <1 , -1> ||)²

= (x - y)² / 2

The locus you are looking for is the same as D²_focus = D²_directrix which is

2[(x - 2)² + (y + 2)²] = (x - y)²

2x² - 8x + 8 + 2y² + 8y + 8 = x² - 2xy + y²

x² + 2xy + y² - 8x + 8y + 16 = 0

Answer 2

You can get it easily by using the distance formula from a point (a, b) to a line Ax+By+C = 0: d = |Aa+Bb+C|/sqrt(A^2+B^2).

d^2 = (x-2)^2 + (y+2)^2 = (x-y)^2 / 2

Simplifying leads to x² + 2xy + y² - 8x + 8y + 16 = 0