Q.If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.

Concept: Relationship between latus rectum, minor axis, and eccentricity of an ellipse.

For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (with $a > b$), the latus rectum is $\ell = \frac{2b^2}{a}$ and the minor axis has length $2b$.

Given that the latus rectum equals half the minor axis:

Simplifying: $\frac{2b^2}{a} = b \implies 2b = a$.

The eccentricity satisfies $b^2 = a^2(1 - e^2)$. Substituting $a = 2b$: …

The latus rectum and the minor axis are both geometric features tied to the ellipse's shape, but they measure different things. The latus rectum is the chord through a focus perpendicular to the major axis, while the minor axis is the shortest diameter. When these satisfy a specific ratio, the ellipse's eccentricity—which quantifies how "stretched" it is—becomes fixed.

For an ellipse with semi-major axis $a$ and semi-minor axis $b$ (where $a > b$), recall that:

• Length of latus rectum: $L = \frac{2b^2}{a}$
• Length of minor axis: $2b$
• Eccentricity: $e = \sqrt{1 - \frac{b^2}{a^2}}$

The problem states that the latus rectum equals half the minor axis. Let's translate that into an equation and extract the eccentricity.

Solution

1. Set up the given condition.

   We're told that the latus rectum equals half the minor axis:

2. Simplify to find the relationship between $a$ and $b$.

   From $\frac{2b^2}{a} = b$, multiply both sides by $a$:

Since $b \neq 0$ (otherwise we wouldn't have an ellipse), divide both sides by $b$:

So the semi-major axis is exactly twice the semi-minor axis: $a = 2b$.

3. Express $\frac{b^2}{a^2}$ in terms of this relationship.

   Substitute $a = 2b$: …