Question

The eccentricity of the ellipse $x^{2}+\frac{y^{2}}{4}=1$ is

• A. $\sqrt{3}$
• B. $\frac{1}{2}$
• C. $\frac{\sqrt{3}}{4}$
• D. $\frac{\sqrt{3}}{2}$
• E. $\frac{1}{\sqrt{3}}$

Hint:

Geometry Tip: Eccentricity is always between 0 and 1 for an ellipse. The formula is always $e = \sqrt{1 - \frac{\text{smaller denominator}}{\text{larger denominator}}}$. This prevents you from ever mixing up $a$ and $b$!

Answer 1

The Correct Option is D

Concept:
The standard equation of an ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The eccentricity $e$ describes how elongated the ellipse is. If $b > a$ (a vertical ellipse), the formula for eccentricity is $e = \sqrt{1 - \frac{a^2}{b^2}}$.

Step 1: Write the equation in explicit standard form.
The given equation is: $$x^2 + \frac{y^2}{4} = 1$$ This can be explicitly written as: $$\frac{x^2}{1} + \frac{y^2}{4} = 1$$

Step 2: Identify the parameters $a^2$ and $b^2$.
By comparing the equation to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we extract: $$a^2 = 1$$ $$b^2 = 4$$

Step 3: Determine the orientation of the ellipse.
Since $b^2 > a^2$ ($4 > 1$), the major axis is along the y-axis. We must use the vertical ellipse eccentricity formula: $$e = \sqrt{1 - \frac{a^2}{b^2}}$$

Step 4: Substitute the values into the formula.
Plug in $a^2 = 1$ and $b^2 = 4$: $$e = \sqrt{1 - \frac{1}{4}}$$

Step 5: Calculate the final value of the eccentricity.
Find the common denominator and simplify the square root: $$e = \sqrt{\frac{4}{4} - \frac{1}{4}}$$ $$e = \sqrt{\frac{3}{4}}$$ $$e = \frac{\sqrt{3}}{2}$$ Hence the correct answer is (D) $\frac{\sqrt{3{2}$}.