Question

The eccentricity of $x^2 + 2y^2 = 3$ is

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

Hint:

Eccentricity for an ellipse always satisfies $0 \lt e \lt 1$. This immediately allows you to eliminate options (B) and (C), as $\sqrt{2} \gt 1$.

Answer 1

The Correct Option is A

To find the eccentricity, we must first transform the given equation into the standard form of an ellipse. Step 1: Standard Form Transformation Divide the entire equation $x^2 + 2y^2 = 3$ by 3 to set the right side to 1: $$\frac{x^2}{3} + \frac{2y^2}{3} = 1 \implies \frac{x^2}{3} + \frac{y^2}{3/2} = 1$$

Step 2: Identify Semi-axes Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $a^2 = 3$ and $b^2 = \frac{3}{2}$. Since $a^2 \gt b^2$, this is a horizontal ellipse where $a$ is the semi-major axis and $b$ is the semi-minor axis.

Step 3: Calculate Eccentricity ($e$) The formula for eccentricity $e$ is: $$e = \sqrt{1 - \frac{b^2}{a^2}}$$ Substituting the values: $$e = \sqrt{1 - \frac{3/2}{3}} = \sqrt{1 - \frac{1}{2}}$$ $$e = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$ Thus, the eccentricity of the given ellipse is $1/\sqrt{2}$.