Question:
<p>The normal at the point (<span class="math-tex">\(b t_{1}^{2}\)</span>, 2bt<sub>1</sub>) on a parabola meets the parabola again in the point (<span class="math-tex">\(b t_{2}^{2}\)</span>, 2bt<sub>2</sub>), then</p>
<p style="display:inline">t<sub>2</sub> = -t<sub>1</sub> <span class="math-tex">\(-\frac{2}{t_{1}}\)</span></p>
<p style="display:inline">t<sub>2</sub> = t<sub>1</sub> <span class="math-tex">\(+\frac{2}{t_{1}}\)</span></p>
<p style="display:inline">t<sub>2</sub> = -t<sub>1</sub> <span class="math-tex">\(+\frac{2}{t_{1}}\)</span></p>
<p style="display:inline">t<sub>2</sub> = t<sub>1</sub> <span class="math-tex">\(-\frac{2}{t_{1}}\)</span></p>
Step-by-Step Solution
Key Concept: The point of intersection where a normal from point t1 meets a parabola again at point t2 is determined by the standard parametric relation t2 = -t1 - 2/t1.
<p>If the normal at the point t<sub>1</sub> meets the parabola y<sup>2</sup> = 4ax again at the point t<sub>2</sub>, then (Standard result)<br />
t<sub>2</sub> = -t<sub>1</sub> <span class="math-tex">$-\frac{2}{t_{1}}$</span>. (Standard result)</p>
Correct Answer: A