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(B) Let the point be $P(h, k)$ and the given points be $A(0, 2)$ and $B(0, -2)$.According to the problem,$PA + PB = 6$.$\sqrt{(h-0)^2 + (k-2)^2} + \sq

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

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(B) Let the point be $P(h, k)$ and the given points be $A(0, 2)$ and $B(0, -2)$.According to the problem,$PA + PB = 6$.$\sqrt{(h-0)^2 + (k-2)^2} + \sqrt{(h-0)^2 + (k+2)^2} = 6$.$\sqrt{h^2 + (k-2)^2} = 6 - \sqrt{h^2 + (k+2)^2}$.Squaring both sides:$h^2 + (k-2)^2 = 36 - 12\sqrt{h^2 + (k+2)^2} + h^2 + (k+2)^2$.$k^2 - 4k + 4 = 36 - 12\sqrt{h^2 + (k+2)^2} + k^2 + 4k + 4$.$-8k - 36 = -12\sqrt{h^2 + (k+2)^2}$.Dividing by $-4$: $2k + 9 = 3\sqrt{h^2 + (k+2)^2}$.Squaring again:$(2k + 9)^2 = 9(h^2 + k^2 + 4k + 4)$.$4k^2 + 36k + 81 = 9h^2 + 9k^2 + 36k + 36$.$9h^2 + 5k^2 = 45$.Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + 5y^2 = 45$.

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

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