The locus of a point P such that PA + PB = 4 | vedclass

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(C) Given points are $A(2, 3, 4)$ and $B(-2, 3, 4)$.The distance $AB = \sqrt{(2 - (-2))^2 + (3 - 3)^2 + (4 - 4)^2} = \sqrt{4^2 + 0 + 0} = 4$.Since $PA

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

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(C) Given points are $A(2, 3, 4)$ and $B(-2, 3, 4)$.The distance $AB = \sqrt{(2 - (-2))^2 + (3 - 3)^2 + (4 - 4)^2} = \sqrt{4^2 + 0 + 0} = 4$.Since $PA + PB = 4$ and $AB = 4$,the point $P$ must lie on the line segment $AB$.For any point $P(x, y, z)$ on the line segment $AB$,the $y$ and $z$ coordinates must be constant,specifically $y = 3$ and $z = 4$.This implies $(y - 3)^2 + (z - 4)^2 = 0$.Expanding this,we get $y^2 - 6y + 9 + z^2 - 8z + 16 = 0$,which simplifies to $y^2 + z^2 - 6y - 8z + 25 = 0$.Thus,the locus is $y^2 + z^2 - 6y - 8z + 25 = 0$ for $x \in [-2, 2]$.

The locus of a point $P$ such that $PA + PB = 4$ where $A(2, 3, 4)$ and $B(-2, 3, 4)$ is

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