Question

The eccentricity of the conic \( x^2 + 2y^2 - 2x + 3y + 2 = 0 \) is

• A. \( 0 \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{1}{2} \)
• D. \( \sqrt{2} \)
• E. \( 1 \)

Hint:

Careful algebra while completing squares is critical for conics.

Answer 1

The Correct Option is B

Concept: For ellipse: \[ e = \sqrt{1 - \frac{b^2}{a^2}} \]

Step 1: Rearrange equation grouping terms.
\[ x^2 - 2x + 2(y^2 + \tfrac{3}{2}y) + 2 = 0 \]

Step 2: Complete square for \( x \).
\[ x^2 - 2x = (x-1)^2 -1 \]

Step 3: Complete square for \( y \).
\[ y^2 + \frac{3}{2}y = \left(y + \frac{3}{4}\right)^2 - \frac{9}{16} \]

Step 4: Substitute back and simplify.
\[ (x-1)^2 -1 + 2\left[\left(y+\frac{3}{4}\right)^2 - \frac{9}{16}\right] +2 =0 \] \[ (x-1)^2 +2\left(y+\frac{3}{4}\right)^2 = \frac{9}{8} \]

Step 5: Convert to standard form.
\[ \frac{(x-1)^2}{9/8} + \frac{(y+3/4)^2}{9/16} = 1 \] \[ a^2 = \frac{9}{8}, \quad b^2 = \frac{9}{16} \]

Step 6: Compute eccentricity.
\[ e = \sqrt{1 - \frac{9/16}{9/8}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}} \]