If A=(2,3,4) and B=(-2,3,4),then the locus of | vedclass

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(C) Let $P = (x, y, z)$. Given $A = (2, 3, 4)$ and $B = (-2, 3, 4)$.$PA + PB = 4$. Since the distance $AB = \sqrt{(-2-2)^2 + (3-3)^2 + (4-4)^2} = \sqr

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$y^2+z^2-6y-8z+25=0$

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(C) Let $P = (x, y, z)$. Given $A = (2, 3, 4)$ and $B = (-2, 3, 4)$.$PA + PB = 4$. Since the distance $AB = \sqrt{(-2-2)^2 + (3-3)^2 + (4-4)^2} = \sqrt{(-4)^2} = 4$,the sum of distances $PA + PB$ is equal to the distance $AB$.This implies that the point $P$ must lie on the line segment $AB$.For any point $P$ on the segment $AB$,the coordinates $y$ and $z$ must be constant,i.e.,$y=3$ and $z=4$.However,the equation $PA+PB=4$ defines a degenerate ellipse (the line segment $AB$).Substituting $y=3$ and $z=4$ into the options,we check for the locus.For option $C$: $(y-3)^2 + (z-4)^2 = 0$,which simplifies to $y^2-6y+9 + z^2-8z+16 = 0$,or $y^2+z^2-6y-8z+25=0$.

If $A=(2,3,4)$ and $B=(-2,3,4)$,then the locus of a point $P(x,y,z)$ such that $PA+PB=4$ is

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