Q56 / 120 · VITEEE 2016 · Mathematics
The normal at the point $\left(\mathrm{at}_{1}^{2}, 2 \mathrm{at}_{1}\right)$ on the parabola, cuts the parabola again at the point whose parameter is
- A$\mathrm{t}_{2}=\mathrm{t}_{1}-\frac{2}{\mathrm{t}_{1}}$
- B$t_{2}=t_{1}+\frac{2}{t_{1}}$
- Correct answerC$t_{2}=-\left(t_{1}+\frac{2}{t_{1}}\right)$
- DNone of these
Solution
Let thenormal a ' $\mathrm{t}_{1}$ ' cuts the parabola again at the point ' $t_{2}$ ' the equation of the normal at $\left(a t_{1}^{2}, 2 a t_{1}\right)$ is $y+t_{1} x=2 a t_{1}+a t_{1}^{3}$
Since it passes through the point ' $\mathrm{t}_{2}$ ' i.e
$$
\begin{array}{l}
\left(\mathrm{at}_{2}^{2}, 2 \mathrm{at}_{2}\right) \\
\therefore \quad 2 \mathrm{at}_{2}+\mathrm{at}_{1} \mathrm{t}_{2}^{2}=2 \mathrm{at}_{1}+\mathrm{at}_{1}^{3} \\
\Rightarrow \quad 2 \mathrm{a}\left(\mathrm{t}_{1}-\mathrm{t}_{2}\right)+\mathrm{at}_{1}\left(\mathrm{t}_{1}^{2}-\mathrm{t}_{2}^{2}\right)=0 \\
\Rightarrow \quad 2+\mathrm{t}_{1}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)=0 \quad\left(\because \mathrm{t}_{1}-\mathrm{t}_{2} \neq 0\right) \\
\Rightarrow \quad 2+\mathrm{t}_{1}^{2}+\mathrm{t}_{1} \mathrm{t}_{2}=0 \\
\Rightarrow \quad \mathrm{t}_{1} \mathrm{t}_{2}=-\left(\mathrm{t}_{1}^{2}+2\right) \Rightarrow \mathrm{t}_{2}=-\left(\mathrm{t}_{1}+\frac{2}{\mathrm{t}_{1}}\right)
\end{array}
$$
Hence, the correct answer is option: 'C'.