Question

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

• A. \(y^2 + z^2 + 6y + 8z + 25 = 0\)
• B. \(y^2 - z^2 + 6y + 8z - 25 = 0\)
• C. \(y^2 + z^2 - 6y - 8z + 25 = 0\)
• D. \(y^2 + z^2 - 6y - 8z - 25 = 0\)

Hint:

When given sum of distances from two symmetric points, use symmetry to reduce dimensions and apply geometric identities.

Answer 1

The Correct Option is C

Let point \(P = (x, y, z)\) Given: \[ PA + PB = 4 \] Where \(A = (2,3,4), B = (-2,3,4)\) \[ PA = \sqrt{(x-2)^2 + (y-3)^2 + (z-4)^2},\quad PB = \sqrt{(x+2)^2 + (y-3)^2 + (z-4)^2} \] Let \(PA + PB = 4\). Squaring both sides doesn't help directly. However, since \(x\)-coordinates of A and B are symmetric about origin and others same, we assume the locus lies in the \(yz\)-plane. So, let \(x = 0\) Now: \[ PA = \sqrt{4 + (y - 3)^2 + (z - 4)^2}
PB = \sqrt{4 + (y - 3)^2 + (z - 4)^2} \Rightarrow PA = PB \] So: \[ PA + PB = 2\sqrt{4 + (y - 3)^2 + (z - 4)^2} = 4 \Rightarrow \sqrt{4 + (y - 3)^2 + (z - 4)^2} = 2 \] Squaring: \[ 4 + (y - 3)^2 + (z - 4)^2 = 4 \Rightarrow (y - 3)^2 + (z - 4)^2 = 0 \Rightarrow y = 3, z = 4 \] This is just a point, but to get the locus, we consider that for symmetric 3D problems of type \(PA + PB = \text{constant}\), the result is an **ellipsoid or a cylinder**, and when simplified here we get: \[ y^2 + z^2 - 6y - 8z + 25 = 0 \]