Find the equation of the normal lines to the curve $3x^{2}-y^{2}=8$ which are parallel to the line $x+3y=4$.

(D) Given equation of the curve is $3x^{2}-y^{2}=8$ $(i)$.
On differentiating both sides with respect to $x$,we get $6x - 2y \frac{dy}{dx} = 0$.
$\Rightarrow \frac{dy}{dx} = \frac{6x}{2y} = \frac{3x}{y}$.
Slope of the tangent $m_{1} = \frac{3x}{y}$.
Slope of the normal $m_{2} = -\frac{1}{m_{1}} = -\frac{y}{3x}$ $(ii)$.
Since the normal is parallel to the line $x+3y=4$,its slope must be equal to the slope of the line.
For the line $x+3y=4$,$y = -\frac{1}{3}x + \frac{4}{3}$,so the slope $m_{3} = -\frac{1}{3}$.
Setting $m_{2} = m_{3}$,we get $-\frac{y}{3x} = -\frac{1}{3} \Rightarrow y = x$ $(iii)$.
Substituting $y=x$ into $(i)$,we get $3x^{2} - x^{2} = 8 \Rightarrow 2x^{2} = 8 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$.
If $x=2$,then $y=2$. If $x=-2$,then $y=-2$. The points are $(2, 2)$ and $(-2, -2)$.
The slope of the normal at these points is $m_{2} = -\frac{y}{3x} = -\frac{2}{3(2)} = -\frac{1}{3}$.
Equation of normal at $(2, 2)$ is $y - 2 = -\frac{1}{3}(x - 2) \Rightarrow 3y - 6 = -x + 2 \Rightarrow x + 3y = 8$.
Equation of normal at $(-2, -2)$ is $y + 2 = -\frac{1}{3}(x + 2) \Rightarrow 3y + 6 = -x - 2 \Rightarrow x + 3y = -8$.
Thus,the required equations are $x + 3y = \pm 8$.