If mx - y + c = 0 is a normal at a point P on | vedclass

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(B) Given,equation of normal $mx - y + c = 0$. Parabola $y^2 = 16x$. Comparing with $y^2 = 4ax$,we get $a = 4$. Let the coordinates of point $P$ be $(

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$132$

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(B) Given,equation of normal $mx - y + c = 0$. Parabola $y^2 = 16x$. Comparing with $y^2 = 4ax$,we get $a = 4$. Let the coordinates of point $P$ be $(at^2, 2at) = (4t^2, 8t)$. The slope of the tangent at $P$ is $\frac{1}{t}$. The slope of the normal is $m = -t$,so $t = -m$. Thus,the point $P$ is $(4m^2, -8m)$. The focal distance of a point $(at^2, 2at)$ is $a(1 + t^2)$. Given focal distance $= 40$,so $4(1 + t^2) = 40$ $\Rightarrow 1 + t^2 = 10$ $\Rightarrow t^2 = 9$ $\Rightarrow t = \pm 3$. Since $t = -m$,$m = \mp 3$,so $m^2 = 9$. The point $P$ is $(4(9), 8(\mp 3)) = (36, \mp 24)$. Since $P$ lies on the normal $mx - y + c = 0$,we have $m(36) - (\mp 24) + c = 0$. Substituting $m = \mp 3$: $(\mp 3)(36) \pm 24 + c = 0 \Rightarrow \mp 108 \pm 24 + c = 0$. If $m = -3$,then $t = 3$,$P = (36, 24)$,normal is $-3x - y + c = 0$ $\Rightarrow -3(36) - 24 + c = 0$ $\Rightarrow -108 - 24 + c = 0$ $\Rightarrow c = 132$. If $m = 3$,then $t = -3$,$P = (36, -24)$,normal is $3x - y + c = 0$ $\Rightarrow 3(36) - (-24) + c = 0$ $\Rightarrow 108 + 24 + c = 0$ $\Rightarrow c = -132$. In both cases,$|c| = 132$.

If $mx - y + c = 0$ is a normal at a point $P$ on the parabola $y^2 = 16x$ and the focal distance of $P$ is $40$ units,then $|c| =$

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