A point moves such that the sum of its distan | vedclass

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Question

(A) Let the point be $P(x, y, z)$. The given points are $A(4, 0, 0)$ and $B(-4, 0, 0)$.According to the problem,$PA + PB = 10$.Using the distance form

Vedclass · Accepted Answer

$9x^2 + 25y^2 + 25z^2 - 225 = 0$

Vedclass · Answer

(A) Let the point be $P(x, y, z)$. The given points are $A(4, 0, 0)$ and $B(-4, 0, 0)$.According to the problem,$PA + PB = 10$.Using the distance formula,$\sqrt{(x-4)^2 + y^2 + z^2} + \sqrt{(x+4)^2 + y^2 + z^2} = 10$.Rearranging,$\sqrt{(x-4)^2 + y^2 + z^2} = 10 - \sqrt{(x+4)^2 + y^2 + z^2}$.Squaring both sides: $(x-4)^2 + y^2 + z^2 = 100 + (x+4)^2 + y^2 + z^2 - 20\sqrt{(x+4)^2 + y^2 + z^2}$.$x^2 - 8x + 16 + y^2 + z^2 = 100 + x^2 + 8x + 16 + y^2 + z^2 - 20\sqrt{(x+4)^2 + y^2 + z^2}$.$-16x - 100 = -20\sqrt{(x+4)^2 + y^2 + z^2}$.Dividing by $-4$: $4x + 25 = 5\sqrt{(x+4)^2 + y^2 + z^2}$.Squaring again: $16x^2 + 200x + 625 = 25(x^2 + 8x + 16 + y^2 + z^2)$.$16x^2 + 200x + 625 = 25x^2 + 200x + 400 + 25y^2 + 25z^2$.$9x^2 + 25y^2 + 25z^2 - 225 = 0$.

$A$ point moves such that the sum of its distances from $(4, 0, 0)$ and $(-4, 0, 0)$ is always $10$. The locus of the point is:

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