Question:
<p>If normal at <span class="math-tex">\(P\left(2, \frac{3 \sqrt{3}}{2}\right)\)</span> meet the major axis of ellipse <span class="math-tex">\(\frac{x^2}{16}+\frac{y^2}{9}=1\)</span> at <span class="math-tex">\(Q\)</span> and <span class="math-tex">\(S^{\prime}\)</span> and <span class="math-tex">\(S\)</span> are foci of given ellipse then SQ: <span class="math-tex">\(S^{\prime} Q\)</span> is</p>
<p style="display:inline"><span class="math-tex">\(\frac{4+\sqrt{7}}{4-\sqrt{7}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{4-\sqrt{7}}{4+\sqrt{7}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{8+\sqrt{7}}{8-\sqrt{7}}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{8-\sqrt{7}}{8+\sqrt{7}}\)</span></p>
Step-by-Step Solution
Key Concept: The normal at point P on an ellipse acts as the internal angle bisector of angle SPS', meaning the ratio SQ:S'Q is equal to the ratio of focal distances SP:S'P by the angle bisector theorem.
<p>Normal at P is the bisector of angle between <span class="math-tex">\({S}^{\prime} {P}\)</span> and SP<br />
Hence <span class="math-tex">\(\frac{{SQ}}{{S}^{\prime} {Q}}=\frac{{SP}}{{S}^{\prime} {P}}=\frac{8-\sqrt{7}}{8+\sqrt{7}}\)</span></p>
Correct Answer: D