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
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
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
D^{2}_{focus} = (x - 2)^{2} + (y + 2)^{2}.
To get the squared distance from the directrix, D^{2}_{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 D^{2}_{directricx} = (<x , y> · <1 , -1> / || <1 , -1> ||)^{2}
= (x - y)^{2} / 2
The locus you are looking for is the same as D^{2}_{focus} = D^{2}_{directrix} which is
2[(x - 2)^{2} + (y + 2)^{2}] = (x - y)^{2}
2x^{2} - 8x + 8 + 2y^{2} + 8y + 8 = x^{2} - 2xy + y^{2}
x^{2} + 2xy + y^{2} - 8x + 8y + 16 = 0
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 x^{2} + 2xy + y^{2} - 8x + 8y + 16 = 0