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 $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

The given ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Since $16 > 9$, the major axis is along the $x$-axis.

Step 1: Identify $a$ and $b$

Here $a^2 = 16 \Rightarrow a = 4$, $b^2 = 9 \Rightarrow b = 3$.

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

For an ellipse, $c^2 = a^2 - b^2 = 16 - 9 = 7 \Rightarrow c = \sqrt{7}$.

Step 3: Compute required quantities

• Vertices: $(\pm a, 0) = (\pm 4, 0)$
• Foci: $(\pm c, 0) = (\pm \sqrt{7}, 0)$
• Length of major axis: $2a = 8$
• Length of minor axis: $2b = 6$
• Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{7}}{4}$
• Length of latus rectum: $\frac{2b^2}{a} = \frac{2 \cdot 9}{4} = \frac{9}{2}$

The equation $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is already in the standard form of an ellipse centered at the origin. The key is to compare it with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since $16 > 9$, the larger denominator is under $x^2$, which tells us the major axis is along the $x$-axis. That means $a^2 = 16$ and $b^2 = 9$, so $a = 4$ and $b = 3$.

For an ellipse, the relationship between $a$, $b$, and the focal distance $c$ is $c^2 = a^2 - b^2$ (since $a > b$). This is the fundamental geometry: the foci lie on the major axis at $(\pm c, 0)$, and the sum of distances from any point on the ellipse to the two foci is constant and equal to $2a$.

Let’s work through each required quantity step by step.

1. Vertices — These are the endpoints of the major axis. Since the major axis is horizontal and centered at the origin, the vertices are at $(\pm a, 0) = (\pm 4, 0)$.

2. Foci — First compute $c$:

   $c^2 = a^2 - b^2 = 16 - 9 = 7$, so $c = \sqrt{7}$.

   The foci are at $(\pm c, 0) = (\pm \sqrt{7}, 0)$.

3. Length of major axis — This is simply $2a = 2 \times 4 = 8$.

4. Length of minor axis — This is $2b = 2 \times 3 = 6$.

5. Eccentricity — Defined as $e = \frac{c}{a}$. So $e = \frac{\sqrt{7}}{4}$.

   Eccentricity tells us how “stretched” the ellipse is; here it’s less than 1, as expected.

6. Length of latus rectum — For an ellipse, the latus rectum is a chord through a focus perpendicular to the major axis. Its length is given by $\frac{2b^2}{a}$.

   So length $= \frac{2 \times 9}{4} = \frac{18}{4} = \frac{9}{2}$.