Find the locus of a point P such that the sum of its distance from ( 0, 2) and ( 0, - 2 ) is 6.

Question

Arun · Answer

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.

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Find the locus of a point P such that the sum of its distance from ( 0, 2) and ( 0, - 2 ) is 6.

Find the locus of a point P such that the sum of its distance from ( 0, 2) and ( 0, - 2 ) is 6.

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