Q.Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 6, 0)$, foci $(\pm 4, 0)$.

Concept: Locus of Point — Ellipse from given vertices and foci.

The vertices are $(\pm 6, 0)$, so the major axis is along the $x$-axis, and $a = 6$.

The foci are $(\pm 4, 0)$, so $c = 4$.

For an ellipse, $c^2 = a^2 - b^2$.

Substitute: $16 = 36 - b^2 \implies b^2 = 20$.

The standard form with centre at origin and major axis horizontal is …

The first thing to notice is what the given points tell us. The vertices are at $(\pm 6, 0)$ — that means the ellipse is stretched farthest along the x‑axis, and the centre is right at the origin. The foci are also on the x‑axis, at $(\pm 4, 0)$. So the major axis is horizontal, and the standard form we need is

where $a$ is the semi‑major axis length (half the total width along the x‑axis) and $b$ is the semi‑minor axis length.

1. Identify $a$ from the vertices.

   The vertices are $(\pm 6, 0)$, so the distance from the centre to a vertex is $a = 6$. That gives $a^2 = 36$.

2. Identify $c$ from the foci.

   The foci are $(\pm 4, 0)$, so the distance from the centre to a focus is $c = 4$. Hence $c^2 = 16$.

3. Use the ellipse relationship to find $b$.

   For any ellipse, the three numbers $a$, $b$, and $c$ are linked by $c^2 = a^2 - b^2$ when the major axis is horizontal. This comes from the geometric definition: the sum of distances from any point on the ellipse to the two foci is constant and equals $2a$.

   So:

Therefore $b = \sqrt{20} = 2\sqrt{5}$, but we only need $b^2$ for the equation. …