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

Tutors, please sign in to answer this question.

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

Vinay J.

Finance, Econ, Math, Stats, Sciences, History, FRM, GMAT and SAT Tutor

New York, NY

4.8
(32 ratings)

Megan L.

Enthusiastic and Dependable Tutor

New York, NY

4.2
(5 ratings)

Christopher S.

Every problem contains its own solution.

Teaneck, NJ