SOLUTION: A point P(x,y) moves so that its distance from the point K(2,5) is twice its distance from the line x=-1. Find the equation of the locus of P.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A point P(x,y) moves so that its distance from the point K(2,5) is twice its distance from the line x=-1. Find the equation of the locus of P.      Log On


   

Question 1132430: A point P(x,y) moves so that its distance from the point K(2,5) is twice its distance from the line x=-1. Find the equation of the locus of P.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
The distance from P(x,y) to the line x=-1 will taken from
the perpendicular distance as shortest which will be a horizontal distance since x=-1 is a vertical line.
This distance will be the difference in the x coordinates
stated as a positive or abs%28x-%28-1%29%29 => abs%28x%2B1%29

Since we want the distance from P(x,y) to K(2,5) to be
twice the distance from (x,y) to x=-1 we have:

sqrt%28%28x-2%29%5E2%2B%28y-5%29%5E2%29+=+2%2Aabs%28x%2B1%29 ... square both sides
Note that since squaring abs%28x%2B1%29 will automatically result in a positive we will no longer need the absolute value sign.

%28x-2%29%5E2%2B%28y-5%29%5E2+=+4%28x%2B1%29%5E2

x%5E2-4x%2B4%2By%5E2-10y%2B25+=+4%28x%5E2%2B2x%2B1%29
x%5E2-4x+%2By%5E2-10y%2B29+=+4x%5E2%2B8x%2B4

0+=+4x%5E2%2B8x%2B4-%28x%5E2-4x+%2By%5E2-10y%2B29%29

+4x%5E2%2B8x%2B4-x%5E2%2B4x+-y%5E2%2B10y-29=0
3x%5E2+%2B+12x+-+y%5E2+%2B+10y+=+25

Note this is becoming the equation of a hyperbola.
We want standard form %28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1+:
We need to do some complete the square work here.

3%28x%5E2%2B4x%29+-+%28y%5E2-10y%29+=+25
3%28x%5E2%2B4x%2B4%29+-+%28y%5E2-10y+%2B+25%29+=+25+%2B+12+-+25

3%28x%2B2%29%5E2+-+%28y-5%29%5E2+=+12

%28x%2B2%29%5E2%2F4+-+%28y-5%29%5E2%2F12+=+1

We have a hyperbola centered at (h,k) or (-2,5).=>the equation of the locus of P