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find the locus of  a point P such that the sum of it's distances from (0,2) and (0,-2)  is 6.
one month ago

Answers : (1)

Arun
22808 Points
							
 
Let point be P(h,k)
((h – 0)^2 + (k – 2)^2)^1/2 + ((h – 0)^2 + (k + 2)^2)^1/2 = 6
(h^2 + (k – 2)^2)^1/2 + (h^2 + (k + 2)^2)^1/2 = 6
(h^2 + (k – 2)^2)^1/2  = 6 – (h^2 + (k + 2)^2)^1/2
now square both sides
h^2 + k^2 – 4k + 4  = 36 + h^2 + k^2 + 4k + 4 – 12(h^2 + (k + 2)^2)^1/2
– 4k  = 36 + 4k – 12(h^2 + (k + 2)^2)^1/2
8k + 36 = 12(h^2 + (k + 2)^2)^1/2
2k + 9 = 3((h^2 + (k + 2)^2)^1/2)
square both sides
4k^2 + 81 + 36k = 9(h^2 + k^2 + 4 + 4k)
9h^2 + 5k^2 = 45
9x^2 + 5y^2 = 45
 
Regards
one month ago
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