If the normal to a parabola y^2 = 4ax at poin | vedclass

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(B) Let the coordinates of $P$ be $(at_1^2, 2at_1)$ and $Q$ be $(at_2^2, 2at_2)$.The slope of the normal at $P(t_1)$ is $m_1 = -t_1 = 	an \alpha$.The

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$-2$

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(B) Let the coordinates of $P$ be $(at_1^2, 2at_1)$ and $Q$ be $(at_2^2, 2at_2)$.The slope of the normal at $P(t_1)$ is $m_1 = -t_1 = 	an \alpha$.The slope of the normal at $Q(t_2)$ is $m_2 = -t_2 = 	an \beta$.The relation between the parameters of the points where the normal at $t_1$ meets the parabola at $t_2$ is $t_2 = -t_1 - \frac{2}{t_1}$.Multiplying by $t_1$,we get $t_1 t_2 = -t_1^2 - 2$,which implies $t_1^2 + t_1 t_2 = -2$.Substituting $t_1 = -	an \alpha$ and $t_2 = -	an \beta$:$(-	an \alpha)^2 + (-	an \alpha)(-	an \beta) = -2$$	an^2 \alpha + 	an \alpha 	an \beta = -2$$	an \alpha (	an \alpha + 	an \beta) = -2$.

If the normal to a parabola $y^2 = 4ax$ at point $P$ meets the curve again at point $Q$,and if $PQ$ and the normal at $Q$ make angles $\alpha$ and $\beta$ respectively with the $x$-axis,then the value of $\tan \alpha (\tan \alpha + \tan \beta)$ is:

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