If the chord joining the points (at_1^2, 2at_ | vedclass

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Question

(A) The equation of the chord passing through $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ is given by $\frac{y - 2at_1}{x - at_1^2} = \frac{2at_2 - 2at_1}

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$t_1t_2 = -1$

Vedclass · Answer

(A) The equation of the chord passing through $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ is given by $\frac{y - 2at_1}{x - at_1^2} = \frac{2at_2 - 2at_1}{at_2^2 - at_1^2} = \frac{2a(t_2 - t_1)}{a(t_2 - t_1)(t_2 + t_1)} = \frac{2}{t_1 + t_2}$.Since the chord passes through the focus $(a, 0)$,we substitute $x = a$ and $y = 0$ into the equation:$\frac{0 - 2at_1}{a - at_1^2} = \frac{2}{t_1 + t_2}$.$\frac{-2at_1}{a(1 - t_1^2)} = \frac{2}{t_1 + t_2}$.$\frac{-t_1}{1 - t_1^2} = \frac{1}{t_1 + t_2}$.$-t_1(t_1 + t_2) = 1 - t_1^2$.$-t_1^2 - t_1t_2 = 1 - t_1^2$.$-t_1t_2 = 1$,which implies $t_1t_2 = -1$.

If the chord joining the points $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ of the parabola $y^2 = 4ax$ passes through the focus of the parabola,then

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