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(D) The equation of the given parabola is $y^2 = 9x$. Here,$4a = 9$,so $a = \frac{9}{4}$.The equation of the normal chord at point $t$ is $y = -tx + 2

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$\pm \sqrt{2}$

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(D) The equation of the given parabola is $y^2 = 9x$. Here,$4a = 9$,so $a = \frac{9}{4}$.The equation of the normal chord at point $t$ is $y = -tx + 2at + at^3$.Substituting $a = \frac{9}{4}$,we get $y = -tx + \frac{9}{2}t + \frac{9}{4}t^3$,which can be written as $tx + y = \frac{9}{2}t + \frac{9}{4}t^3$.Since the chord subtends a right angle at the vertex $V(0,0)$,we homogenize the parabola $y^2 = 9x$ using the line equation:$y^2 = 9x \left( \frac{tx + y}{\frac{9}{2}t + \frac{9}{4}t^3} ight)$$y^2 = 9x \left( \frac{tx + y}{\frac{9}{4}t(2 + t^2)} ight) = \frac{4x(tx + y)}{t(2 + t^2)}$$t(2 + t^2)y^2 = 4tx^2 + 4xy$$4tx^2 + 4xy - t(2 + t^2)y^2 = 0$For a right angle at the origin,the sum of coefficients of $x^2$ and $y^2$ must be zero:$4t - t(2 + t^2) = 0$Since $t 
eq 0$,we divide by $t$:$4 - (2 + t^2) = 0$$4 - 2 - t^2 = 0$$t^2 = 2 \Rightarrow t = \pm \sqrt{2}$.Thus,option $D$ is correct.

If a normal chord at a point $t (\neq 0)$ on the parabola $y^2 = 9x$ subtends a right angle at its vertex,then $t =$

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