Question

Find the locus of a point p such that the sum of its distances from

Answer 1

Step-by-step explanation:

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

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