If a normal is drawn to the parabola y^2 = 4a | vedclass

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(D) For the parabola $y^2 = 4ax$,the coordinates of a point on the parabola are given by $(at^2, 2at)$.Comparing $(2a, 2\sqrt{2}a)$ with $(at^2, 2at)$

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$6\sqrt{3}a$

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(D) For the parabola $y^2 = 4ax$,the coordinates of a point on the parabola are given by $(at^2, 2at)$.Comparing $(2a, 2\sqrt{2}a)$ with $(at^2, 2at)$,we get $t^2 = 2$,so $t = \sqrt{2}$.The length of the normal chord at parameter $t$ is given by the formula $L = 2a|t| (t^2 + 2)$.Substituting $t = \sqrt{2}$:$L = 2a(\sqrt{2})((\sqrt{2})^2 + 2)$$L = 2a(\sqrt{2})(2 + 2)$$L = 2a(\sqrt{2})(4) = 8\sqrt{2}a$.Wait,re-evaluating the standard formula for the length of a normal chord $L = 4a \frac{(1+m^2)^{3/2}}{m^2}$ where $m = -t$ is the slope of the normal.For $t = \sqrt{2}$,the slope of the normal $m = -t = -\sqrt{2}$.$L = 4a \frac{(1 + (-\sqrt{2})^2)^{3/2}}{(-\sqrt{2})^2} = 4a \frac{(1+2)^{3/2}}{2} = 2a(3)^{3/2} = 2a(3\sqrt{3}) = 6\sqrt{3}a$.

If a normal is drawn to the parabola $y^2 = 4ax$ at the point $(2a, 2\sqrt{2}a)$,what is the length of the normal chord?

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