If the line x-y=-4K is a tangent to the parab | vedclass

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(C) The equation of the line is $y = x + 4K$. For this to be a tangent to the parabola $y^2 = 8x$ (where $a=2$),the condition $c = a/m$ must be satisf

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$\frac{9}{2\sqrt{2}}$

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(C) The equation of the line is $y = x + 4K$. For this to be a tangent to the parabola $y^2 = 8x$ (where $a=2$),the condition $c = a/m$ must be satisfied.Here,$c = 4K$,$a = 2$,and $m = 1$.So,$4K = 2/1 \implies 4K = 2 \implies K = 1/2$.The point of contact $P$ is given by $(a/m^2, 2a/m) = (2/1^2, 2(2)/1) = (2, 4)$.The equation of the normal to the parabola $y^2 = 4ax$ at $(x_1, y_1)$ is $y - y_1 = -\frac{y_1}{2a}(x - x_1)$.Substituting $x_1 = 2, y_1 = 4, a = 2$: $y - 4 = -\frac{4}{2(2)}(x - 2) \implies y - 4 = -1(x - 2) \implies x + y - 6 = 0$.The point $(K, 2K)$ is $(1/2, 1)$.The perpendicular distance from $(1/2, 1)$ to $x + y - 6 = 0$ is $d = \frac{|1/2 + 1 - 6|}{\sqrt{1^2 + 1^2}} = \frac{|3/2 - 6|}{\sqrt{2}} = \frac{|-9/2|}{\sqrt{2}} = \frac{9}{2\sqrt{2}}$.

If the line $x-y=-4K$ is a tangent to the parabola $y^2=8x$ at $P$,then the perpendicular distance of the normal at $P$ from $(K, 2K)$ is

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