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

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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

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