The locus of a point such that the sum of its | vedclass

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(C) Let $P(x, y)$ be any point on the locus. According to the problem,the sum of distances from $P$ to $F_1(0, 2)$ and $F_2(0, -2)$ is $6$. $\sqrt{x^2

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$9x^2 + 5y^2 = 45$

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(C) Let $P(x, y)$ be any point on the locus. According to the problem,the sum of distances from $P$ to $F_1(0, 2)$ and $F_2(0, -2)$ is $6$. $\sqrt{x^2 + (y - 2)^2} + \sqrt{x^2 + (y + 2)^2} = 6$ $\sqrt{x^2 + (y - 2)^2} = 6 - \sqrt{x^2 + (y + 2)^2}$ Squaring both sides: $x^2 + y^2 - 4y + 4 = 36 + x^2 + y^2 + 4y + 4 - 12\sqrt{x^2 + (y + 2)^2}$ $-8y - 36 = -12\sqrt{x^2 + (y + 2)^2}$ $2y + 9 = 3\sqrt{x^2 + (y + 2)^2}$ Squaring again: $4y^2 + 36y + 81 = 9(x^2 + y^2 + 4y + 4)$ $4y^2 + 36y + 81 = 9x^2 + 9y^2 + 36y + 36$ $9x^2 + 5y^2 = 45$

The locus of a point such that the sum of its distances from the points $(0, 2)$ and $(0, -2)$ is $6$ is:

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