Parabola
11.0 Relation between parametric coefficients if normal meets parabola
11.0 Relation between parametric coefficients if normal meets parabola
Relation between $t_1$ and $t_2$ if normal at $t_1$ meets parabola again at $t_2$.
Let the equation of parabola be ${y^2} = 4ax$ and the equation of normal at $P\left( {a{t_1}^2,2a{t_1}} \right)$ is $$y = - {t_1}x + 2a{t_1} + a{t_1}^3...(1)$$
Since it meets the parabola again at $Q(a{t_2}^2,2a{t_2})$ which satisfies the equation $(1)$
Therefore, $$2a{t_2} = - {t_1}\left( {a{t_2}^2} \right) + 2a{t_1} + a{t_1}^3$$ $$2a{t_2} = - a{t_1}{t_2}^2 + 2a{t_1} + a{t_1}^3$$ $$2a\left( {{t_2} - {t_1}} \right) + a{t_1}\left( {{t_2}^2 - {t_1}^2} \right) = 0$$ $$a\left( {{t_2} - {t_1}} \right)\left[ {2 + {t_1}\left( {{t_2} + {t_1}} \right)} \right] = 0$$
Since, ${t_1}$ and ${t_2}$ are different
$$a\left( {{t_2} - {t_1}} \right) \ne 0$$
Therefore, $$2 + {t_1}\left( {{t_2} + {t_1}} \right) = 0$$ or, $${t_2} = - {t_1} - \frac{2}{{{t_1}}}$$
Note: If normal at ${t_1}$ and ${t_2}$ meets the parabola ${y^2} = 4ax$ at same point then ${t_1} \times {t_2} = 2$.
Proof: Let us assume that the normal meets at point having parameter $T$.
Therefore, $$T = - {t_1} - \frac{2}{{{t_1}}} = - {t_2} - \frac{2}{{{t_2}}}$$
or, $${t_1} - {t_2} = 2\left( {\frac{1}{{{t_1}}} - \frac{1}{{{t_2}}}} \right)$$ $$\left( {{t_1} - {t_2}} \right)\left[ {1 - \frac{2}{{{t_1}{t_2}}}} \right] = 0$$
Since, ${t_1}$ and ${t_2}$ are different $$\left( {{t_2} - {t_1}} \right) \ne 0$$
Therefore, $$\left[ {1 - \frac{2}{{{t_1}{t_2}}}} \right] = 0$$ or, $${t_1}{t_2} = 2$$
