Q.Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$.

The given ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$. Since $36 > 16$, the major axis is along the $x$-axis. This is the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 = 36$ and $b^2 = 16$, so $a = 6$, $b = 4$.

Step 1: Find $c$ (distance from centre to focus)

For an ellipse, $c^2 = a^2 - b^2 = 36 - 16 = 20$, so $c = \sqrt{20} = 2\sqrt{5}$.

Step 2: Vertices and foci

Vertices are at $(\pm a, 0) = (\pm 6, 0)$.

Foci are at $(\pm c, 0) = (\pm 2\sqrt{5}, 0)$.

Step 3: Lengths and eccentricity

Length of major axis $= 2a = 12$.

Length of minor axis $= 2b = 8$.

Eccentricity $e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.

Length of latus rectum $= \frac{2b^2}{a} = \frac{2 \cdot 16}{6} = \frac{32}{6} = \frac{16}{3}$.

The standard form of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the semi-major axis and $b$ is the semi-minor axis. The larger denominator tells you which axis is longer. Here, $36 > 16$, so $a^2 = 36$ and $b^2 = 16$, meaning the major axis lies along the $x$-axis. This choice determines everything else: foci lie on the major axis, vertices are at the ends of the major axis, and the eccentricity measures how "stretched" the ellipse is.

Let’s extract the numbers step by step.

1. Identify $a$ and $b$

   $a^2 = 36 \implies a = 6$ (semi-major axis)

   $b^2 = 16 \implies b = 4$ (semi-minor axis)

   Since $a > b$, the major axis is horizontal.

2. Find $c$ (distance from center to each focus)

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

   $c^2 = 36 - 16 = 20 \implies c = \sqrt{20} = 2\sqrt{5}$.

   The foci are on the major axis, so coordinates: $(\pm c, 0) = (\pm 2\sqrt{5}, 0)$.

3. Vertices

   These are the endpoints of the major axis: $(\pm a, 0) = (\pm 6, 0)$.

4. Lengths of axes

   Major axis length = $2a = 12$

   Minor axis length = $2b = 8$

5. Eccentricity

   $e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.

   Since $e < 1$, it’s an ellipse (a circle would have $e=0$).

6. Length of latus rectum

   The latus rectum is a chord through a focus perpendicular to the major axis. Its length is $\frac{2b^2}{a}$.

   $\frac{2 \cdot 16}{6} = \frac{32}{6} = \frac{16}{3}$.