Question:
<p>The length of the minor axis (along the y-axis) of an ellipse in the standard form is <span class="math-tex">\(\frac{4}{\sqrt{3}}\)</span>. If this ellipse touches the line, x + 6y = 8; then its eccentricity is:</p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2} \sqrt{\frac{5}{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{2} \sqrt{\frac{11}{3}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\sqrt{\frac{5}{6}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{1}{3} \sqrt{\frac{11}{3}}\)</span></p>
Step-by-Step Solution
Key Concept: Use the minor axis length to determine b and the condition of tangency c^2 = a^2m^2 + b^2 to find a^2, then calculate the eccentricity.
<p>Let <span class="math-tex">\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 ; a>b\)</span><br />
<span class="math-tex">\(2 b=\frac{4}{\sqrt{3}} \Rightarrow b=\frac{2}{\sqrt{3}}\)</span><br />
Equation of tangent <span class="math-tex">\(\equiv y=m x \pm \sqrt{a^{2} m^{2}+b^{2}}\)</span><br />
Comparing with <span class="math-tex">\(\equiv y=\frac{-x}{6}+\frac{4}{3}\)</span><br />
<span class="math-tex">\(m=\frac{-1}{6}\)</span> and a<sup>2</sup>m<sup>2</sup> + b<sup>2</sup> <span class="math-tex">\(=\frac{16}{9}\)</span><br />
<span class="math-tex">\(\Rightarrow \frac{a^{2}}{36}+\frac{4}{3}=\frac{16}{9} \Rightarrow \frac{a^{2}}{36}=\frac{16}{9}-\frac{4}{3}=\frac{4}{9}\)</span><br />
<span class="math-tex">\(\Rightarrow\)</span> a<sup>2</sup> = 16 <span class="math-tex">\(\Rightarrow\)</span> a = <span class="math-tex">\(\pm\)</span>4<br />
Now, the eccentricity of the ellipse <span class="math-tex">\((e)=\sqrt{1-\frac{b^{2}}{a^{2}}}\)</span><br />
<span class="math-tex">\(\Rightarrow e=\sqrt{1-\frac{4}{3 \times 16}}=\sqrt{\frac{11}{12}}=\frac{1}{2} \sqrt{\frac{11}{3}}\)</span></p>
Correct Answer: B