Question

find the locus of P(x,y) such that the sum of the square of its distance from (-1,0) and (1,0)

Answer 1

the sentence is imcoplete.

Answer 2

*incomplete

Answer 3

is 26 sorry

Answer 4

|(x-1)^2+y^2|+|(x+1)^2+y^2\=26?

Answer 5

let A(-1,0) and B(1,0) and P(x,y). if i understand it is
\[ \large [d(P,A)+d(P,B)]^2=26 \]
am i wrong?

Answer 6

isnt it ((d(P,A))^2 + other one?

Answer 7

i don't know. the wording of the problem is confusing

Answer 8

sum so + of square of dist idkk

Answer 9

yes... it should be: \(\large d(P,A)^2+d(P,B)^2=26 \)

Answer 10

so just absolute valcue?

Answer 11

hyelpp??

Answer 12

well if @dpaInc is correct, then using the definition of distance
\[ \large 26=d(P,A)^2+d(P,B)^2= \]
\[ \large =[(x+1)^2+(y-0)^2]+[(x-1)^2+(y-0)^2] \]

Answer 13

okay so just do it normally??

Answer 14

yes