Statement (A): If the normal at the ends of t | vedclass

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(A) For the parabola $y^2 = 4x$,we have $a = 1$.The ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(1, 2)$ and $(1, -2)$.For $y^2 =

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$A$ and $R$ are both independently true and $R$ is the correct explanation for $A$.

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(A) For the parabola $y^2 = 4x$,we have $a = 1$.The ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$,which are $(1, 2)$ and $(1, -2)$.For $y^2 = 4ax$,the parameter $T$ at $(at^2, 2at)$ is $t$. Here $t^2 = 1$,so $T_1 = \pm 1$.The relation for the normal at $T_1$ meeting the parabola again at $T_2$ is $T_2 = -T_1 - \frac{2}{T_1}$.For $T_1 = 1$,$T_2 = -1 - 2 = -3$. The point is $(aT_2^2, 2aT_2) = (9, -6)$.For $T_1 = -1$,$T_2 = 1 + 2 = 3$. The point is $(aT_2^2, 2aT_2) = (9, 6)$.The distance $PP' = \sqrt{(9-9)^2 + (6 - (-6))^2} = \sqrt{0^2 + 12^2} = 12$ units.Both $A$ and $R$ are true,and $R$ explains $A$.

Statement $(A)$: If the normal at the ends of the latus rectum of the parabola $y^2 = 4x$ meet the curve again at $P$ and $P'$,then $PP' = 12$ units.
Reason $(R)$: If the normal at $T_1$ to the parabola $y^2 = 4ax$ meets the parabola again at $T_2$,then $T_2 = -T_1 - \frac{2}{T_1}$.

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Statement $(A)$: If the normal at the ends of the latus rectum of the parabola $y^2 = 4x$ meet the curve again at $P$ and $P'$,then $PP' = 12$ units.Reason $(R)$: If the normal at $T_1$ to the parabola $y^2 = 4ax$ meets the parabola again at $T_2$,then $T_2 = -T_1 - \frac{2}{T_1}$.