If the normal at the point (bt_1^2, 2bt_1) on | vedclass

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(A) The equation of the parabola is $y^2 = 4bx$.Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4b$,so $\frac{dy}{dx} = \frac{2b}{y}$.A

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$t_2 = -t_1 - \frac{2}{t_1}$

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(A) The equation of the parabola is $y^2 = 4bx$.Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4b$,so $\frac{dy}{dx} = \frac{2b}{y}$.At the point $(bt_1^2, 2bt_1)$,the slope of the tangent is $\frac{2b}{2bt_1} = \frac{1}{t_1}$.Therefore,the slope of the normal at $(bt_1^2, 2bt_1)$ is $m = -t_1$.The equation of the normal at $(bt_1^2, 2bt_1)$ is $(y - 2bt_1) = -t_1(x - bt_1^2)$.Since the point $(bt_2^2, 2bt_2)$ lies on this normal,we substitute these coordinates into the equation:$(2bt_2 - 2bt_1) = -t_1(bt_2^2 - bt_1^2)$$2b(t_2 - t_1) = -t_1b(t_2 - t_1)(t_2 + t_1)$Dividing by $b(t_2 - t_1)$ (assuming $t_2 
eq t_1$ and $b 
eq 0$):$2 = -t_1(t_2 + t_1)$$2 = -t_1t_2 - t_1^2$$t_1t_2 = -t_1^2 - 2$$t_2 = -t_1 - \frac{2}{t_1}$

If the normal at the point $(bt_1^2, 2bt_1)$ on the parabola $y^2 = 4bx$ meets the parabola again at the point $(bt_2^2, 2bt_2)$,then:

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