Question:
<p>If y = x and 3y + 2x = 0 are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is</p>
<p style="display:inline"><span class="math-tex">\(\frac{2}{\sqrt{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\sqrt{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{\sqrt{2}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\sqrt{\frac{2}{3}}\)</span></p>
Step-by-Step Solution
Key Concept: The eccentricity of an ellipse is found by equating the product of the slopes of conjugate diameters to -b²/a² and substituting the axis ratio into the eccentricity formula b²/a² = 1 - e².
<p>Let the equation of the ellipse be <span class="math-tex">$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$</span><br />
Now, slopes of the given diameters are m<sub>1</sub> = 1, m<sub>2</sub> = <span class="math-tex">$\frac{-2}{3}$</span><br />
<span class="math-tex">$\Rightarrow$</span> m<sub>1</sub>m<sub>2</sub> <span class="math-tex">$=\frac{-2}{3}=\frac{-b^{2}}{a^{2}}$</span><br />
<span class="math-tex">$\Rightarrow$</span> 3b<sup>2</sup> = 2a<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> 3a<sup>2</sup>(1 - e<sup>2</sup>) = 2a<sup>2</sup><br />
<span class="math-tex">$\Rightarrow$</span> 1 - e<sup>2</sup> = <span class="math-tex">$\frac{2}{3}$</span><br />
<span class="math-tex">$\Rightarrow e^{2}=\frac{1}{3}$</span><br />
<span class="math-tex">$\Rightarrow e=\frac{1}{\sqrt{3}}$</span></p>
Correct Answer: B