The normal at a point on the parabola y^2 = 4 | vedclass

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(C) Let the parabola be $y^2 = 4ax$ with $a = 1$. The equation of a normal to the parabola $y^2 = 4x$ at point $(t^2, 2t)$ is $y = -tx + 2t + t^3$.If

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$5$

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(C) Let the parabola be $y^2 = 4ax$ with $a = 1$. The equation of a normal to the parabola $y^2 = 4x$ at point $(t^2, 2t)$ is $y = -tx + 2t + t^3$.If this normal passes through $P(h, k)$,then $k = -th + 2t + t^3$,which simplifies to $t^3 + (2-h)t - k = 0$.Let the roots of this cubic equation be $t_1, t_2, t_3$. These are the parameters of the feet of the three normals.The coordinates of the feet are $(t_1^2, 2t_1), (t_2^2, 2t_2), (t_3^2, 2t_3)$.The centroid $G(x_g, y_g)$ of the triangle formed by these points is given by $x_g = \frac{t_1^2 + t_2^2 + t_3^2}{3}$ and $y_g = \frac{2(t_1 + t_2 + t_3)}{3}$.From the cubic equation $t^3 + (2-h)t - k = 0$,we have $\sum t_i = 0$ and $\sum t_i t_j = 2-h$.Since $\sum t_i = 0$,the $y$-coordinate of the centroid is $y_g = 0$,which matches $G(2,0)$.Now,$x_g = \frac{(\sum t_i)^2 - 2\sum t_i t_j}{3} = \frac{0^2 - 2(2-h)}{3} = \frac{2(h-2)}{3}$.Given $x_g = 2$,we have $\frac{2(h-2)}{3} = 2$,which implies $h-2 = 3$,so $h = 5$.

The normal at a point on the parabola $y^2 = 4x$ passes through a point $P$. Two more normals to this parabola also pass through $P$. If the centroid of the triangle formed by the feet of these three normals is $G(2,0)$,then the abscissa of $P$ is

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