Question:
<p>Consider the ellipse E: <span class="math-tex">\(\frac{x^{2}}{16}+\frac{v^{2}}{9}=1\)</span>, with the major axis AA'. P is a point of E and Q is the corresponding point on the major auxiliary circle. M is the midpoint of PQ. The eccentricity of locus of M:</p>
<p style="display:inline">is <span class="math-tex">\(\frac{\sqrt{15}}{8}\)</span></p>
<p style="display:inline">is <span class="math-tex">\(\sqrt 2\)</span></p>
<p style="display:inline">does not exist</p>
<p style="display:inline">is 1</p>
Step-by-Step Solution
Key Concept: Use parametric coordinates (a cos θ, b sin θ) and (a cos θ, a sin θ) for the ellipse and its major auxiliary circle to derive the locus of the midpoint.
<p>P = (4 cos <span class="math-tex">\(\theta\)</span>, 3 sin<span class="math-tex">\(\theta\)</span>) and Q = (4 cos<span class="math-tex">\(\theta\)</span>, 4 sin<span class="math-tex">\(\theta\)</span>)<br />
<span class="math-tex">\(\Rightarrow\)</span> M = <span class="math-tex">\(\left(4 \cos \theta, \frac{7}{2} \sin \theta\right)\)</span> = (h, k), say<br />
Then <span class="math-tex">\(\left(\frac{h}{4}\right)^{2}+\left(\frac{k}{7 / 2}\right)^{2}=1\)</span> (an ellipse)<br />
Let e be the eccentricity.<br />
Then <span class="math-tex">\(\left(\frac{7}{2}\right)^{2}\)</span> = (4)<sup>2</sup> (1 - e<sup>2</sup>)<br />
<span class="math-tex">\(\Rightarrow\)</span> e<sup>2</sup> <span class="math-tex">\(=1-\frac{49}{4 \times 16}\)</span><br />
<span class="math-tex">\(\Leftrightarrow e^{2}=\frac{15}{64} \Rightarrow e=\frac{\sqrt{15}}{8}\)</span></p>
Correct Answer: A