Question

Eccentricity of the ellipse \( 4x^2 + y^2 - 8x + 4y - 8 = 0 \) is

• A. \( \frac{\sqrt{3}}{2} \)
• B. \( \frac{\sqrt{3}}{4} \)
• C. \( \frac{\sqrt{3}}{\sqrt{2}} \)
• D. \( \frac{\sqrt{3}}{8} \)
• E. \( \frac{\sqrt{3}}{16} \)

Hint:

Always identify larger denominator as \( a^2 \) when finding eccentricity.

Answer 1

The Correct Option is A

Concept: Standard ellipse: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] Eccentricity: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Step 1: Rearrange equation.
\[ 4x^2 -8x + y^2 +4y -8 = 0 \]

Step 2: Complete squares.
\[ 4(x^2 -2x) + (y^2 +4y) = 8 \] \[ 4[(x-1)^2 -1] + [(y+2)^2 -4] = 8 \]

Step 3: Simplify.
\[ 4(x-1)^2 -4 + (y+2)^2 -4 = 8 \] \[ 4(x-1)^2 + (y+2)^2 = 16 \]

Step 4: Divide by 16.
\[ \frac{(x-1)^2}{4} + \frac{(y+2)^2}{16} = 1 \] So: \[ a^2 = 16,\quad b^2 = 4 \]

Step 5: Compute eccentricity.
\[ e = \sqrt{1 - \frac{4}{16}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]