Question:
<p>If the normal to the ellipse 3x<sup>2</sup> + 4y<sup>2</sup> = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to</p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{61}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{221}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{5 \sqrt{5}}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{\sqrt{157}}{2}\)</span></p>
Step-by-Step Solution
Key Concept: Use the parametric form of the ellipse to relate the slope of the normal to the given line and solve for the coordinates of point P using the tangent's intersection with Q.
<p>Equation of given ellipse is 3x<sup>2</sup> + 4y<sup>2</sup> = 12<br />
<span class="math-tex">\(\Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{3}=1\)</span> ...(i)<br />
Now, let point P(2 cos<span class="math-tex">\(\theta\)</span>, <span class="math-tex">\(\sqrt3\)</span> sin<span class="math-tex">\(\theta\)</span>), so equation of tangent to ellipse (i) at point P is<br />
<span class="math-tex">\(\frac{x \cos \theta}{2}+\frac{y \sin \theta}{\sqrt{3}}=1\)</span> ...(ii)<br />
Since, tangent (ii) passes through point Q(4, 4)<br />
<span class="math-tex">\(\therefore 2 \cos \theta+\frac{4}{\sqrt{3}} \sin \theta=1\)</span> ...(iii)<br />
and equation of normal to ellipse (i) at point P is<br />
<span class="math-tex">\(\frac{4 x}{2 \cos \theta}-\frac{3y}{\sqrt{3} \sin \theta}=4-3\)</span><br />
<span class="math-tex">\(\Rightarrow\)</span> <span class="math-tex">\(2 x \sin \theta-\sqrt{3} \cos \theta y=\sin \theta \cos \theta\)</span> ...(iv)<br />
Since, normal (iv) is parallel to line, 2x + y = 4<br />
<span class="math-tex">\(\therefore\)</span> Slope of normal (iv) is parallel to line, 2x + y = 4<br />
<span class="math-tex">\(\Rightarrow \quad \frac{2}{\sqrt{3}} \tan \theta=-2 \Rightarrow \tan \theta=-\sqrt{3} \Rightarrow \theta=120^{\circ}\)</span><br />
<span class="math-tex">\(\Rightarrow(\sin \theta, \cos \theta)=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\)</span><br />
Hence, point <span class="math-tex">\(P\left(-1, \frac{3}{2}\right)\)</span><br />
Now, PQ = <span class="math-tex">\(=\sqrt{(4+1)^{2}+\left(4-\frac{3}{2}\right)^{2}}\)</span> [given cordinates of Q <span class="math-tex">\(\equiv\)</span> (4, 4)]<br />
<span class="math-tex">\(=\sqrt{25+\frac{25}{4}}=\frac{5 \sqrt{5}}{2}\)</span></p>
Correct Answer: C