Question:
<p>The eccentricity of the curve represented by the equation x<sup>2</sup> + 2y<sup>2</sup> - 2x + 3y + 2 = 0 is</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\sqrt{2}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\sqrt 2\)</span></p>
<p style="display:inline">0</p>
Step-by-Step Solution
Key Concept: Transform the general quadratic equation into the standard form of an ellipse by completing the square for both variables to identify the lengths of the semi-axes.
<p>Given equation of ellipse is<br />
x<sup>2</sup> + 2y<sup>2</sup> - 2x + 3y + 2 = 0<br />
<span class="math-tex">$\Rightarrow \frac{(x-1)^{2}}{2}+\left(y+\frac{3}{4}\right)^{2}=\frac{1}{16}$</span><br />
<span class="math-tex">$\Rightarrow \frac{(x-1)^{2}}{\left(\frac{1}{8}\right)}+\frac{\left(y+\frac{3}{4}\right)^{2}}{\left(\frac{1}{16}\right)}=1$</span><br />
which is an ellipse with <span class="math-tex">$a^{2}=\frac{1}{8}$</span> and <span class="math-tex">$b^{2}=\frac{1}{16}$</span><br />
Since, a > b<br />
<span class="math-tex">$e^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{1}{2} \Rightarrow e=\frac{1}{\sqrt{2}}$</span></p>
Correct Answer: B