Question

The eccentricity of the ellipse \[ \frac{(x - 1)^2}{2} + \left(y + \frac{3}{4}\right)^2 = \frac{1}{16} \] is: 

• A. \( \frac{1}{\sqrt{2}} \)
• B. \( \frac{1}{2\sqrt{2}} \)
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{4} \)
• E. \( \frac{1}{4\sqrt{2}} \)

Hint:

Eccentricity is always between 0 and 1 for an ellipse. If your calculation yields something else, check if you swapped \( a^2 \) and \( b^2 \). The denominator of the smaller term always goes in the numerator of the fraction inside the root.

Answer 1

The Correct Option is A

Concept: The eccentricity \( e \) of an ellipse measures how "stretched" it is. For an ellipse in standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), the eccentricity is calculated as \( e = \sqrt{1 - \frac{\text{minor axis}^2}{\text{major axis}^2}} \). We must first bring the given equation into this standard form.

Step 1: Convert the equation to standard form.
Divide the entire equation by \( \frac{1}{16} \): \[ \frac{\frac{(x-1)^2}{2}}{\frac{1}{16}} + \frac{(y + \frac{3}{4})^2}{\frac{1}{16}} = 1 \] \[ \frac{(x-1)^2}{2 \cdot \frac{1}{16}} + \frac{(y + \frac{3}{4})^2}{\frac{1}{16}} = 1 \] \[ \frac{(x-1)^2}{1/8} + \frac{(y + \frac{3}{4})^2}{1/16} = 1 \]

Step 2: Identify \( a^2 \) and \( b^2 \).
Comparing with the standard form:
• \( a^2 = \frac{1}{8} \)
• \( b^2 = \frac{1}{16} \) Since \( \frac{1}{8} > \frac{1}{16} \), the major axis is along the \( x \)-direction.

Step 3: Calculate the eccentricity \( e \).
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1/16}{1/8}} \] \[ e = \sqrt{1 - \frac{1}{16} \times \frac{8}{1}} = \sqrt{1 - \frac{1}{2}} \] \[ e = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]