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# If ${t_1}$ and ${t_2}$ are two extremities of any focal chord of the parabola ${y^2} = 4ax$ then${t_1}{t_2}=$A. $- 1$B. $0$C. $\pm 1$D. $\dfrac{1}{2}$

Last updated date: 19th Mar 2023
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Coordinates of end point of focal chord are $\left( {at_1^2,2a{t_1}} \right),\left( {at_2^2,2a{t_2}} \right)$and the focus is$\left( {a,0} \right)$
$\Rightarrow \dfrac{{2a{t_2} - 2a{t_1}}}{{at_2^2 - at_1^2}} = \dfrac{{2a{t_2} - 0}}{{at_2^2 - a}} \\ \Rightarrow \dfrac{{{t_2} - {t_1}}}{{t_2^2 - t_1^2}} = \dfrac{{{t_2}}}{{t_2^2 - 1}} \\ \Rightarrow \dfrac{{{t_2} - {t_1}}}{{\left( {{t_2} - {t_1}} \right)\left( {{t_2} + {t_1}} \right)}} = \dfrac{{{t_2}}}{{t_2^2 - 1}} \\ \Rightarrow \dfrac{1}{{\left( {{t_2} + {t_1}} \right)}} = \dfrac{{{t_2}}}{{t_2^2 - 1}} \\ \Rightarrow t_2^2 - 1 = {t_2}\left( {{t_2} + {t_1}} \right) = {t_1}{t_2} + t_2^2 \\ \Rightarrow {t_1}{t_2} = - 1 \\$