If y = mx + c is the normal at a point on the | vedclass

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(C) For the parabola $y^2 = 4ax$, we have $4a = 8$, so $a = 2$.The focal distance of a point $(at^2, 2at)$ on the parabola is given by $a(1 + t^2)$.Gi

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$10$

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(C) For the parabola $y^2 = 4ax$, we have $4a = 8$, so $a = 2$.The focal distance of a point $(at^2, 2at)$ on the parabola is given by $a(1 + t^2)$.Given the focal distance is $8$, we have $2(1 + t^2) = 8$, which implies $1 + t^2 = 4$, so $t^2 = 3$ and $t = \pm\sqrt{3}$.The equation of the normal to the parabola $y^2 = 4ax$ at parameter $t$ is $y = -tx + 2at + at^3$.Comparing this with $y = mx + c$, we get $m = -t$ and $c = 2at + at^3 = at(2 + t^2)$.Substituting $a = 2$ and $t = \sqrt{3}$ (or $t = -\sqrt{3}$ for the magnitude):$|c| = |2(\sqrt{3})(2 + 3)| = |2\sqrt{3}(5)| = 10\sqrt{3}$.

If $y = mx + c$ is the normal at a point on the parabola $y^2 = 8x$ whose focal distance is $8 \text{ units}$, then $|c|$ is equal to (in $\sqrt{3}$)

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