Question

If the eccentricity of the ellipse \( ax^2 + 4y^2 = 4a \, (a<4) \) is \( \frac{1}{\sqrt{2}} \), then its semi-minor axis is equal to:

• A. $2$
• B. $\sqrt{2}$
• C. $1$
• D. $\sqrt{3}$
• E. $3$

Hint:

The semi-minor axis is the square root of the smaller denominator in the standard form of the ellipse. Always pay attention to constraints like $(a < 4)$ as they dictate which axis is the major one.

Answer 1

The Correct Option is B

Concept: Standardize the ellipse equation to identify $a^2$ and $b^2$. The eccentricity $e$ for an ellipse is given by $b^2 = a^2(1 - e^2)$ if $a > b$, or $a^2 = b^2(1 - e^2)$ if $b > a$.

Step 1: Normalize the equation.
Divide $ax^2 + 4y^2 = 4a$ by $4a$: \[ \frac{ax^2}{4a} + \frac{4y^2}{4a} = 1 \quad \Rightarrow \quad \frac{x^2}{4} + \frac{y^2}{a} = 1 \] Here, $X^2$ denominator is $4$ and $Y^2$ denominator is $a$. We are given $a < 4$, so the major axis is along the x-axis ($a_{major}^2 = 4$) and the minor axis is along the y-axis ($b_{minor}^2 = a$).

Step 2: Use the eccentricity to find $a$.
Given $e = \frac{1}{\sqrt{2}}$. Using $b^2 = a^2(1 - e^2)$: \[ a = 4 \left( 1 - \left(\frac{1}{\sqrt{2}}\right)^2 \right) = 4 \left( 1 - \frac{1}{2} \right) = 4 \times \frac{1}{2} = 2 \] So, the square of the semi-minor axis is $b^2 = a = 2$.

Step 3: Find the semi-minor axis.
The semi-minor axis is $b = \sqrt{a}$: \[ b = \sqrt{2} \]