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 $\frac{x^2}{25} + \frac{y^2}{100} = 1$.

The given ellipse is $\frac{x^2}{25} + \frac{y^2}{100} = 1$. Since $100 > 25$, the major axis is along the $y$-axis.

Step 1 – Identify $a$ and $b$

Here $a^2 = 100 \Rightarrow a = 10$ (semi-major axis), $b^2 = 25 \Rightarrow b = 5$ (semi-minor axis).

Step 2 – Find $c$ (distance from centre to focus)

For an ellipse, $c^2 = a^2 - b^2 = 100 - 25 = 75 \Rightarrow c = \sqrt{75} = 5\sqrt{3}$.

Step 3 – Write all required values

• Vertices: $(0, \pm a) = (0, \pm 10)$
• Foci: $(0, \pm c) = (0, \pm 5\sqrt{3})$
• Length of major axis: $2a = 20$
• Length of minor axis: $2b = 10$ …

1. Recognising the standard form

The given equation is

For an ellipse centred at the origin, the standard form is

where $a$ is the semi‑major axis and $b$ the semi‑minor axis — but only if $a > b$. The larger denominator tells us which axis is the major axis.

Here $25 < 100$, so the bigger number is under $y^2$. That means the major axis is vertical (along the y‑axis). We must therefore assign:

2. Vertices

For a vertical ellipse centred at $(0,0)$, the vertices lie on the y‑axis at a distance $a$ from the centre:

3. Foci

The foci are also on the major axis, inside the ellipse. Their distance from the centre is $c$, where

Substitute:

Since the major axis is vertical, the foci are at

4. Lengths of major and minor axes

• Major axis length = $2a = 2 \times 10 = 20$.
• Minor axis length = $2b = 2 \times 5 = 10$.

5. Eccentricity

Eccentricity $e$ measures how “stretched” the ellipse is:

$e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2}.$ …