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

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(D) Given the parabola equation $x^2 = 4y$ and the chord equation $x - \sqrt{2}y + 4\sqrt{2} = 0$.From the chord equation,we have $y = \frac{x + 4\sqr

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$6\sqrt{3}$

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(D) Given the parabola equation $x^2 = 4y$ and the chord equation $x - \sqrt{2}y + 4\sqrt{2} = 0$.From the chord equation,we have $y = \frac{x + 4\sqrt{2}}{\sqrt{2}} = \frac{x}{\sqrt{2}} + 4$.Substitute $y$ into the parabola equation: $x^2 = 4(\frac{x}{\sqrt{2}} + 4) = 2\sqrt{2}x + 16$.Rearranging gives the quadratic equation: $x^2 - 2\sqrt{2}x - 16 = 0$.Let the roots be $x_1$ and $x_2$. Then $x_1 + x_2 = 2\sqrt{2}$ and $x_1x_2 = -16$.The difference of the roots is $|x_1 - x_2| = \sqrt{(x_1 + x_2)^2 - 4x_1x_2} = \sqrt{(2\sqrt{2})^2 - 4(-16)} = \sqrt{8 + 64} = \sqrt{72} = 6\sqrt{2}$.Since $y = \frac{x}{\sqrt{2}} + 4$,the difference in $y$-coordinates is $|y_1 - y_2| = |\frac{x_1 - x_2}{\sqrt{2}}| = \frac{6\sqrt{2}}{\sqrt{2}} = 6$.The length of the chord is $\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = \sqrt{(6\sqrt{2})^2 + 6^2} = \sqrt{72 + 36} = \sqrt{108} = 6\sqrt{3}$.

The length of the chord of the parabola $x^2 = 4y$ having the equation $x - \sqrt{2}y + 4\sqrt{2} = 0$ is

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