Q.Find the equation of the set of points $P$, the sum of whose distances from $A(4, 0, 0)$ and $B(-4, 0, 0)$ is equal to $10$.

Concept: Locus of a point — we find the set of all points $P(x, y, z)$ satisfying a given distance condition.

Let $P = (x, y, z)$. The given condition is:

where $PA = \sqrt{(x-4)^2 + y^2 + z^2}$ and $PB = \sqrt{(x+4)^2 + y^2 + z^2}$.

Step 1: Write the equation:

Step 2: Square both sides after isolating one radical, then simplify. A cleaner approach: note this is the definition of an ellipsoid of revolution (prolate spheroid) with foci at $A$ and $B$, and constant sum $2a = 10$, so $a = 5$. Distance between foci $2c = 8$, so $c = 4$. Then $b^2 = a^2 - c^2 = 25 - 16 = 9$. …

The problem asks for the locus of a point $P(x,y,z)$ such that $PA + PB = 10$, with $A(4,0,0)$ and $B(-4,0,0)$. This is the 3D version of the classic ellipse definition: the set of points whose sum of distances to two fixed points (foci) is constant. In space, that surface is an ellipsoid of revolution (a prolate spheroid) with the line joining the foci as its axis of symmetry.

Why does this work? The distance sum condition forces $P$ to lie on a surface where the $x$-axis is the major axis, and the $y$ and $z$ directions are symmetric — the shape is the same in any plane containing the $x$-axis. So we can first solve in the $xy$-plane (a 2D ellipse) and then extend to 3D by replacing $y^2$ with $y^2+z^2$.

Let’s go step by step.

1. Set up the distance condition. Let $P = (x, y, z)$. Then

The condition is

2. Isolate one square root and square. Move one term to the other side:

Square both sides:

3. Simplify the equation. Cancel $y^2+z^2$ on both sides. Expand the squares:

Cancel $x^2$ and $16$ from both sides:

Bring terms together:

Multiply by $-1$:

4. Divide and square again. Divide by 4:

Square both sides:

Expand:

5. Cancel and rearrange. The $200x$ terms cancel. Bring everything to one side:

Multiply by $-1$:

$9x^2 + 25y^2 + 25z^2 = 225.$ …