Question:
<p>The normal to the curve y<sup>2</sup> = 4ax at P(t<sub>1</sub>) meets the curve again at Q(t<sub>2</sub>). If <span class="math-tex">\(\angle \mathrm{POQ}=\frac{\pi}{2}\)</span>, where O is the origin, then</p>
<p style="display:inline"><span class="math-tex">\(t_{2}^{2}\)</span> = 2</p>
<p style="display:inline">t<sub>1</sub> = 2t<sub>2</sub></p>
<p style="display:inline"><span class="math-tex">\(t_{1}^{2}\)</span> = 2</p>
<p style="display:inline">t<sub>2</sub> = 2t<sub>1</sub></p>
Step-by-Step Solution
Key Concept: Utilize the specific parametric relationship for a normal chord of a parabola and combine it with the perpendicularity condition for slopes of lines connecting the origin to points on the curve.
<p>The normal at P(t<sub>1</sub>) meets the curve again at Q(t<sub>2</sub>)<br />
<span class="math-tex">\(\Rightarrow\)</span> t<sub>2</sub> = t<sub>1</sub> - <span class="math-tex">\(\frac{2}{t_{1}}\)</span> ...(i)<br />
Slope of OP <span class="math-tex">\(=\frac{2 a t_{1}}{a t_{1}^{2}}=\frac{2}{t_{1}}\)</span><br />
Similarly slope of OQ = <span class="math-tex">\(\frac{2}{t_{2}}\)</span><br />
OP <span class="math-tex">\(\perp\)</span> OQ <span class="math-tex">\(\Rightarrow\left(\frac{2}{t_{1}}\right)\left(\frac{2}{t_{2}}\right)=-1\)</span><br />
<span class="math-tex">\(\Leftrightarrow\)</span> t<sub>1</sub>t<sub>2</sub> = -4 ...(ii)<br />
(i) and (ii) give<br />
t<sub>1</sub>t<sub>2</sub> = <span class="math-tex">\(-t_{1}^{2}\)</span> - 2<br />
<span class="math-tex">\(\Leftrightarrow t_{1}^{2}\)</span> = -2 + 4 = 2</p>
Correct Answer: C