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(B) Let the point be $P(x, y, z)$. The given points are $F_1(a, 0, 0)$ and $F_2(-a, 0, 0)$.According to the problem,$PF_1 + PF_2 = 2k$.Using the dista

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$\frac{x^2}{k^2}+\frac{y^2+z^2}{k^2-a^2}=1$

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(B) Let the point be $P(x, y, z)$. The given points are $F_1(a, 0, 0)$ and $F_2(-a, 0, 0)$.According to the problem,$PF_1 + PF_2 = 2k$.Using the distance formula: $\sqrt{(x-a)^2 + y^2 + z^2} + \sqrt{(x+a)^2 + y^2 + z^2} = 2k$.Let $d^2 = y^2 + z^2$. Then $\sqrt{(x-a)^2 + d^2} = 2k - \sqrt{(x+a)^2 + d^2}$.Squaring both sides: $(x-a)^2 + d^2 = 4k^2 + (x+a)^2 + d^2 - 4k\sqrt{(x+a)^2 + d^2}$.$x^2 - 2ax + a^2 = 4k^2 + x^2 + 2ax + a^2 - 4k\sqrt{(x+a)^2 + d^2}$.$-4ax - 4k^2 = -4k\sqrt{(x+a)^2 + d^2}$.$ax + k^2 = k\sqrt{(x+a)^2 + d^2}$.Squaring again: $a^2x^2 + 2axk^2 + k^4 = k^2(x^2 + 2ax + a^2 + y^2 + z^2)$.$a^2x^2 + 2axk^2 + k^4 = k^2x^2 + 2axk^2 + k^2a^2 + k^2(y^2 + z^2)$.$x^2(a^2 - k^2) + k^2(y^2 + z^2) = k^2a^2 - k^4 = -k^2(k^2 - a^2)$.Dividing by $-k^2(k^2 - a^2)$: $\frac{x^2(a^2 - k^2)}{-k^2(k^2 - a^2)} + \frac{k^2(y^2 + z^2)}{-k^2(k^2 - a^2)} = 1$.$\frac{x^2}{k^2} + \frac{y^2 + z^2}{k^2 - a^2} = 1$.

For $a \neq 0$,if the sum of the distances of a point $P(x, y, z)$ from the points $F_1(a, 0, 0)$ and $F_2(-a, 0, 0)$ is a constant $2k$,then the locus of that point is

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