$\begin{array}{1 1}(A)\;(-18,-12) \\(B)\;(-18,12) \\(C)\;(18,12) \\(D)\;(18,-12) \end{array}$

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$y^2=8x$ -----(1)

=> $ 2y. \large\frac{dy}{dx}$$=8$

Slope of tangent at $(2,4)= \bigg[\large\frac{dy}{dx}\bigg]_{(2,4)}$

$\qquad=1$

Slope of normal at $(2.4)=-1$

Equation of normal at $(2,4)$ is

$x+y-6=0$ ----(2)

Solving (1) and (2) we get,

$x=2;y=4$

and $x=18,y=-12$

$\therefore $ other point is $(18,-12)$

Hence D is the correct answer.

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