find the equation ofthe locus of a point which moves in the plane so that its distance from (-2,0) is always three times its distance fromt he origin
thats a parabola
the question after it says shows that the locus is a circle...
the distance from its single foci is 3times its distance to the origin .... might be eliptic
i cant tell :)
hey amistre.
yes? :)
if v hav a graph of x = 1, then wht will b y ?
y will be lonely
but i cant write this LONELY thing,
\(\color{#f2e526}{\text{i}}\)\(\color{#3f5611}{\text{f}}\)\(\color{#b55857}{\text{ }}\)\(\color{#52da8d}{\text{x}}\)\(\color{#43eb8c}{\text{=}}\)\(\color{#6ed573}{\text{1}}\)\(\color{#2fcdae}{\text{;}}\)\(\color{#2dadc9}{\text{ }}\)\(\color{#a2bf6f}{\text{y}}\)\(\color{#3bddb5}{\text{ }}\)\(\color{#fb28fd}{\text{w}}\)\(\color{#35b173}{\text{i}}\)\(\color{#da931a}{\text{l}}\)\(\color{#e1c579}{\text{l}}\)\(\color{#d2396b}{\text{ }}\)\(\color{#5398c5}{\text{a}}\)\(\color{#eefeac}{\text{l}}\)\(\color{#3adb86}{\text{w}}\)\(\color{#aabcb4}{\text{a}}\)\(\color{#d9ac7d}{\text{y}}\)\(\color{#bd46ed}{\text{s}}\)\(\color{#fc8c8d}{\text{ }}\)\(\color{#1c55a2}{\text{b}}\)\(\color{#ffa143}{\text{e}}\)\(\color{#1de66e}{\text{ }}\)\(\color{#bbf359}{\text{l}}\)\(\color{#d5f6ff}{\text{o}}\)\(\color{#cedad1}{\text{n}}\)\(\color{#4c94be}{\text{e}}\)\(\color{#f7ad29}{\text{l}}\)\(\color{#49a7f2}{\text{y}}\)\(\color{#864fb7}{\text{}}\)
lol okay...
y not exist on a vertical line do it?
no.
umm so...circle...or no...where'd amistre go?
but wht should i write?
location of a point which moves in the plane its distance from (-2,0) is always 3x distance from (0,0)
yes...
what is its distance if it IS the origin
[\sqrt{(x-0)^2+(y-0)^2}]
for my final answer i got 2x^2-4x-4
the point (-3,0) is on this path; and so is (-1.5 , 0 )
right?
right!
and we did that because thats our new second point?
thats just one point that is 3 times the distance to the origin as it is to the -2,0
i just tried to figure it for a point above and below the point; but my angle comes back as 0 or 180
this question is very interesting...
its cause I tried to force it to be 3 and 1 lol instead of a ration
if i did it right; we get points above and below it at a distance of 1/sqrt(2)
well the good news is it aint a circle
thats bad news lol...
lets flip it over to the positive side from 0 to 2 is our origin and point; 3,0 is 1 away from 2 and 3 away fromm 0; its good 1.5 is 3/2 from 0 and 1/2 from 2 so we have 2 points (1.5,0) and (3,0) that satisfy our misinterpretations
u are over complicating it
the center of these 2 points is of course 2.25; if this was a circle then the point that is above and below it would match our problem
overcomplicating is what I do when im using what i know to get to where i dont know
\[\sqrt{(x+2)^2 +y^2} = 3 \sqrt{ x^2+y^2}\]
square both sides expand, group and complete the squares
.75 would be our radius by my conjecture; so the point (2.25, .75) would have to pythag to = 9
(x^2 +4x +4 +y^2 = 9 ( x^2 +y^2)
well, not 9; but 3(.75 ) to be good
8x^2 +8y^2 -4x -4 =0
divide through by 4
no 8
x^2 -(1/2)x +y^2 = (1/2)
(x-(1/4) )^2 +y^2 = (1/2) + (1/16)
(x-(1/4))^2 +y^2 = 9/16
i got the same thing but i forgot tot square my 3
circle ecentre ( (1/4) , 0 ) radius 3/4
if hes right; id go with it cause I cant get nothing beating my head against the wall lol
lol aww thanks so much amistre...
so elece you used method of completing of the squares to find the centre and radius
THANKS SO MUCH ELECENGINEER!!! and AMISTRE!!!
9/16 is not 3/4
circle is x^2 +y^2 = r^2
sqrt(9/16)= 3/4
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