Analytical Geometry> find the locus of a point p such that the...

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

Aryan Todi , 8 Years ago

Grade 11

2 Answers

Ritesh Khatri

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

Last Activity: 8 Years ago

Ashwin Sheoran

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!!!

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Last Activity: 7 Years ago