The length of the chord of the parabola y^2 = | vedclass

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(D) The equation of the chord of the parabola $y^2 = 4ax$ bisected at $(x_1, y_1)$ is given by $T = S_1$. Here,$y^2 = x$,so $4a = 1 \Rightarrow a = 1/

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$2\sqrt{5}$

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(D) The equation of the chord of the parabola $y^2 = 4ax$ bisected at $(x_1, y_1)$ is given by $T = S_1$. Here,$y^2 = x$,so $4a = 1 \Rightarrow a = 1/4$. The equation of the chord is $y y_1 - 2a(x + x_1) = y_1^2 - 4ax_1$. Substituting $(x_1, y_1) = (2, 1)$ and $a = 1/4$: $y(1) - 2(1/4)(x + 2) = 1^2 - 4(1/4)(2)$ $y - 1/2(x + 2) = 1 - 2$ $y - 1/2x - 1 = -1$ $y = 1/2x \Rightarrow x = 2y$. Substitute $x = 2y$ into $y^2 = x$: $y^2 = 2y \Rightarrow y(y - 2) = 0$. So,$y = 0$ or $y = 2$. If $y = 0$,$x = 0$. Point $A = (0, 0)$. If $y = 2$,$x = 4$. Point $B = (4, 2)$. The length of the chord $AB = \sqrt{(4 - 0)^2 + (2 - 0)^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.

The length of the chord of the parabola $y^2 = x$ which is bisected at the point $(2, 1)$ is

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