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

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3 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)^1/2now square both sidesh^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)^1/28k + 36 = 12(h^2 + (k + 2)^2)^1/22k + 9 = 3((h^2 + (k + 2)^2)^1/2)square both sides4k^2 + 81 + 36k = 9(h^2 + k^2 + 4 + 4k)9h^2 + 5k^2 = 459x^2 + 5y^2 = 45
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3 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 ·
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2 years ago
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