Question:
<p>If the eccentricity of an ellipse be <span class="math-tex">\(\frac 58\)</span> and the distance between its foci be 10, then its latus rectum is:</p>
<p style="display:inline">12</p>
<p style="display:inline"><span class="math-tex">\(\frac {37}2\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac {39}4\)</span></p>
<p style="display:inline">15</p>
Step-by-Step Solution
Key Concept: Determine the semi-axes $a$ and $b$ using the focal distance ($2c$) and eccentricity ($e = c/a$) to compute the length of the latus rectum using the formula $2b^2/a$.
<p>According to the given, 2c = 10 <span class="math-tex">$\Rightarrow$</span> <span class="math-tex">$\frac{10}{2}$</span> = 5<br />
We know that eccentricity = <span class="math-tex">$\frac{\mathbf{c}}{\mathbf{a}}$</span><br />
<span class="math-tex">$\Rightarrow$</span> <span class="math-tex">$\frac{5}{8}=\frac{5}{a}$</span> <span class="math-tex">$\Rightarrow$</span> a = 8<br />
Also, <span class="math-tex">$c^{2}=a^{2}-b^{2}$</span><br />
<span class="math-tex">$\Rightarrow $</span> b<sup>2 </sup>= a<sup>2 </sup>- c<sup>2 </sup>= 8<sup>2 </sup>- 5<sup>2 </sup>= 64 - 25 = 39<br />
Length of the latus-rectum = <span class="math-tex">$\frac{2 b^{2}}{a}=\frac{2 \times 39}{8}=\frac{39}{4}$</span></p>
Correct Answer: C