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 $\frac{x^2}{100} + \frac{y^2}{400} = 1$.

The key idea is that the ellipse is in standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, but here the larger denominator is under $y^2$, so the major axis is vertical.

Step 1 – Identify $a$ and $b$

Since $400 > 100$, we have $b^2 = 400$ and $a^2 = 100$.

Thus $b = 20$ (semi-major axis), $a = 10$ (semi-minor axis).

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

For an ellipse, $c^2 = b^2 - a^2 = 400 - 100 = 300$, so $c = \sqrt{300} = 10\sqrt{3}$.

Step 3 – Write all required values

• Vertices: $(0, \pm b) = (0, \pm 20)$
• Foci: $(0, \pm c) = (0, \pm 10\sqrt{3})$
• Length of major axis: $2b = 40$
• Length of minor axis: $2a = 20$ …

1. Identify the standard form and orientation

The given equation is

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

where $a$ is the semi-major axis length if $a > b$ (horizontal ellipse), and $b$ is the semi-major axis length if $b > a$ (vertical ellipse).

Here $a^2 = 100$ so $a = 10$, and $b^2 = 400$ so $b = 20$. Since $b > a$, the major axis is along the y-axis. That means the ellipse is taller than it is wide.

2. Vertices

For a vertical ellipse centred at $(0,0)$, the vertices lie on the y-axis at $(0, \pm b)$.

So $b = 20$ gives vertices at

The length of the major axis is $2b = 40$.

3. Minor axis

The minor axis is along the x-axis, with semi-minor axis $a = 10$.

So the endpoints of the minor axis are $(\pm 10, 0)$, and its total length is $2a = 20$.

4. Eccentricity

For an ellipse, eccentricity $e$ is given by

when $b > a$ (vertical major axis).

Substitute $a^2 = 100$, $b^2 = 400$:

5. Foci

For a vertical ellipse, the foci lie on the major axis (y-axis) at $(0, \pm c)$, where

So

Thus the foci are at …