Q.Find the equation of the ellipse whose vertices are $(\pm 13, 0)$ and foci are $(\pm 5, 0)$.

Concept: Standard form of an ellipse with horizontal major axis.

The vertices $(\pm 13, 0)$ lie on the $x$-axis, so the major axis is horizontal. For such an ellipse centered at the origin, the standard form is

where $a$ is the semi-major axis and $b$ is the semi-minor axis.

Step 1: From the vertices $(\pm 13, 0)$, we have $a = 13$.

Step 2: From the foci $(\pm 5, 0)$, we have $c = 5$. …

An ellipse is the set of all points whose distances from two fixed points (the foci) sum to a constant. When the ellipse is centered at the origin with its major axis along the $x$-axis, the standard form is

where $a > b > 0$. The vertices are the farthest points on the ellipse, located at $(\pm a, 0)$, and the foci are at $(\pm c, 0)$ where $c^2 = a^2 - b^2$.

The geometry here is straightforward: the vertices tell us how far the ellipse stretches horizontally, the foci tell us how "squashed" it is, and the relationship $c^2 = a^2 - b^2$ connects these to the vertical semi-axis $b$.

Finding the equation

1. Identify $a$ from the vertices

The vertices are at $(\pm 13, 0)$, so the semi-major axis is $a = 13$. This gives us $a^2 = 169$.

2. Identify $c$ from the foci

The foci are at $(\pm 5, 0)$, so $c = 5$, which means $c^2 = 25$.

3. Calculate $b^2$ using the fundamental relationship

For any ellipse, the relationship between $a$, $b$, and $c$ is:

Substituting our values:

So $b = 12$. …