# find the locus of a point p such that the sum of its distances from (0 2) and (0 -2) is 6

Ritesh Khatri
76 Points
5 years ago
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

Ashwin Sheoran
44 Points
4 years ago
By distance formula,
[x^2 + (y-2)^2]^1/2 + [x^2 + (y+2)^2]^1/2 = 6.
(x^2 + y^2 - 4y + 4)^1/2 + (x^2 + y^2 + 4y + 4)^1/2 = 6.
Let x^2 + y^2 + 4 = a.
(a - 4y)^1/2 + (a + 4y)^1/2 = 6.
(a - 4y)^1/2 = 6 - (a + 4y)^1/2.
Squaring both sides,
a - 4y = a + 4y + 36 - 12* (a + 4y)^1/2.
8y + 36 = 12* (a + 4y)^1/2.
2y + 9 = 3* (a + 4y)^1/2.
Squaring again,
4y^2 + 36y + 81 = 9a + 36y.
9a - 4y^2 - 81 = 0.
Putting back the original value of a,
9x^2 + 5y^2 - 45 = 0.
And we're done!!!
7.4k Views ·