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

Question

anonymous · Answer

Anyone?

anonymous · Answer

use the distance formula

anonymous · Answer

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

anonymous · Answer

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

dumbcow · Answer

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

anonymous · Answer

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

dumbcow · Answer

nevermind im wrong

anonymous · Answer

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

dumbcow · Answer

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

dumbcow · Answer

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