$\begin{array}{1 1}(a)\;(9,-6)\\(b)\;(8,-6)\\(c)\;(7,-6)\\(d)\;(9,6)\end{array}$

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We know normal to the parabola $y^2=4x$ is

$y-y_1=-\large\frac{-y_1}{2a}$$(x-x_1)$

$(y-2)=-\large\frac{2}{2\times 1}$$(x-1)$

$y+x=3$

$y^2=4x$

$y^2=4(3-y)$

$y^2+4y-12=0$

$y^2-2y+6y-12=0$

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

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

$y=2,-6$

$3-y=x$

$3+6=x$

$x=9$

Hence (9,-6) is another point on the parabola.

Hence (a) is the correct answer.

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