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 $\frac{x^2}{49} + \frac{y^2}{36} = 1$.

Concept: Standard form of an ellipse

The equation $\frac{x^2}{49} + \frac{y^2}{36} = 1$ is in standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 = 49$ and $b^2 = 36$.

Since $a^2 > b^2$, the major axis lies along the $x$-axis. We have $a = 7$ and $b = 6$.

The relationship $b^2 = a^2(1 - e^2)$ gives $36 = 49(1 - e^2)$, so $e^2 = \frac{13}{49}$ and $e = \frac{\sqrt{13}}{7}$.

The distance from center to focus is $c = ae = 7 \cdot \frac{\sqrt{13}}{7} = \sqrt{13}$.

Results:

• Foci: $(\pm\sqrt{13}, 0)$
• Vertices: $(\pm 7, 0)$
• Length of major axis: $2a = 14$
• Length of minor axis: $2b = 12$ …

Understanding the Standard Form

An ellipse in standard position centered at the origin has the equation

The larger denominator tells us which axis is major. Here $49 > 36$, so the major axis lies along the $x$-axis. The relationship $a^2 = b^2 + c^2$ (where $c$ is the distance from center to each focus) connects the semi-axes to the focal distance. Every other property—vertices, foci, eccentricity, latus rectum—flows from $a$, $b$, and $c$.

Step-by-Step Analysis

1. Identify $a^2$ and $b^2$

Comparing with the standard form:

Since $a^2 > b^2$, the major axis is horizontal.

2. Calculate $a$ and $b$

3. Find $c$ using the fundamental relation

4. Locate the vertices

The vertices lie at the ends of the major axis, which extends $a$ units from the origin along the $x$-axis:

5. Locate the foci

The foci lie on the major axis, $c$ units from the center:

6. Determine the length of the major axis

The major axis spans from one vertex to the other:

7. Determine the length of the minor axis

The minor axis extends $b$ units above and below the center along the $y$-axis:

8. Calculate the eccentricity

Eccentricity measures how "stretched" the ellipse is: