Question

find the locus of a point p(x,y) which moves so that its distance from the point (1,0) is half its distance from the line x=-1

Answer 1

Anyone?

Answer 2

use the distance formula

Answer 3

\[\sqrt{(x-1)^2+(y-0)^2}=\frac{1}{2}\sqrt{(x+1)^2+(y-y)^2}\]

Answer 4

no 2x^2-4x+2+2y^2=x^2+2x+1

Answer 5

i get an ellipse, gonna double check it
\[\frac{6(x-\frac{5}{3})^{2}}{19} + \frac{8y^{2}}{19} = 1\]

Answer 6

x^2+2y^2-2x+1=0

Answer 7

nevermind im wrong

Answer 8

Thanks alot though!...
And Purplec16 Thanks! got that too i jus wanted to double check.

Answer 9

hmm purple you have to square the 1/2 as well when you square both sides
4y^2 +4x^2 -8x +4 = x^2 +2x +1

Answer 10

found my mistake, its half of an ellipse
\[\frac{9(x-\frac{5}{3})^{2}}{16} + \frac{3y^{2}}{4} = 1\]
for x<= 5/3