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(C) The given equation of the pair of lines is $2x^2+4xy-4y^2-6x-8y+7=0$.To find the length of the intercept on the $Y$-axis,we set $x=0$.Substituting

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$\sqrt{11}$

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(C) The given equation of the pair of lines is $2x^2+4xy-4y^2-6x-8y+7=0$.To find the length of the intercept on the $Y$-axis,we set $x=0$.Substituting $x=0$ into the equation,we get $-4y^2-8y+7=0$,which is equivalent to $4y^2+8y-7=0$.Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we have $y = \frac{-8 \pm \sqrt{64 - 4(4)(-7)}}{2(4)} = \frac{-8 \pm \sqrt{64+112}}{8} = \frac{-8 \pm \sqrt{176}}{8}$.This simplifies to $y = \frac{-8 \pm 4\sqrt{11}}{8} = -1 \pm \frac{\sqrt{11}}{2}$.The two points of intersection on the $Y$-axis are $y_1 = -1 + \frac{\sqrt{11}}{2}$ and $y_2 = -1 - \frac{\sqrt{11}}{2}$.The length of the intercept is $|y_1 - y_2| = |(-1 + \frac{\sqrt{11}}{2}) - (-1 - \frac{\sqrt{11}}{2})| = |\sqrt{11}| = \sqrt{11}$.

Find the length of the intercept cut by the pair of lines $2x^2+4xy-4y^2-6x-8y+7=0$ on the $Y$-axis.

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