Question

Find the equation of the locus of a point p which moves such that the sum of the squares of its distances from (3,0) and (-3,0) is 68

Answer 1

@rational

Answer 2

hi!!

Answer 3

HI!!

Answer 4

i guess we can do this, it is gong to take a bit of algebra

Answer 5

ok

Answer 6

lets call such a point \((x,y)\) and make a equation

Answer 7

ok

Answer 8

the square of the distance between \((x,y)\) and \((3,0)\) is \((x-3)^2+y^2\)

Answer 9

the square of the distance between \((x,y)\) and \((-3,0)\) is \[(x+3)^2+y^2\]

Answer 10

add them together, set the result equal to \(68\) get \[(x-3)^2+(x+3)^2+2y^2=68\]
now more algebra

Answer 11

ok so x^2-6x+9+y^2
and x^2+6x+9+y^2
so adding them together we get 2x^2+18+2y^2
like that?

Answer 12

that is what i get, yes

Answer 13

oh equal 68

Answer 14

\[2x^2+18+2y^2 =68\] subtract \(18\) maybe to get \[2x^2+2y^2=50\] then divide by \(2\) to make it look neater

Answer 15

\[x^2+y^2=25\] might might be a nicer answer
a circle with center \((0,0)\) and radius \(5\)

Answer 16

so thats the equation of the locus

Answer 17

??

Answer 18

yes

Answer 19

assuming we did all the algebra right
looks good to me

Answer 20

but we didnt show that thee distance is 68..how would we do that?

Answer 21

the sum of the square of the distances was supposed to be 68 right?

Answer 22

yea

Answer 23

\[(x-3)^2+(x+3)^2+2y^2\] was the sum of the square of the distances
\[(x-3)^2+(x+3)^2+2y^2=68\] says it is equal to 68

Answer 24

ok

Answer 25

then the algebra turned that in to \[x^2+y^2=25\]

Answer 26

ok i understand now thank you

Answer 27

but ...what happen to the square root sign

Answer 28

\[(\sqrt{(x-3)^2+(y-0)^2})^2+(\sqrt{(x+3)^2+(y-0)^2})^2=68\]ok i see now

Answer 29

yea thank you !!

Answer 30

lol \[\color\magenta\heartsuit\]