Q.Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$, foci $(\pm 4, 0)$.

Concept: Locus of Point — The ellipse is defined by the sum of distances to two fixed foci being constant. Here, the vertices and foci lie on the $x$-axis, so the major axis is horizontal.

Step 1: For vertices $(\pm a, 0)$, we have $a = 5$. For foci $(\pm c, 0)$, we have $c = 4$.

Step 2: The relationship for an ellipse is $c^2 = a^2 - b^2$. Substituting:

$16 = 25 - b^2 \implies b^2 = 9$, so $b = 3$. …

The first thing to notice is where the vertices and foci are placed. Both pairs are symmetric about the origin and lie on the x‑axis: $(\pm 5, 0)$ and $(\pm 4, 0)$. That tells you immediately that the centre of the ellipse is at $(0,0)$ and the major axis is horizontal. For an ellipse, the vertices are at $(\pm a, 0)$ when the major axis is along the x‑axis, so $a = 5$. The foci are at $(\pm c, 0)$, so $c = 4$.

The fundamental relationship for any ellipse is $c^2 = a^2 - b^2$, where $b$ is the semi‑minor axis length. This comes from the definition: the sum of distances from any point on the ellipse to the two foci is constant and equals $2a$. If you work that geometry out at the topmost point $(0, b)$, you get the same relation.

So we compute $b^2$:

1. Identify $a$ and $c$ from the given points.

   Vertices $(\pm 5, 0)$ mean $a = 5$.

   Foci $(\pm 4, 0)$ mean $c = 4$.

2. Apply the ellipse relation.

Hence $b = 3$ (the positive root, since it’s a length).

3. Write the standard equation. For a horizontal major axis centred at the origin, the equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$ …