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

Given equation of the curve is
$3{x^2} - {y^2} = 8$ …..(i)

On differentiating both sides w.r.t. $x$, we get
$6x - 2y\frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{6x}}{{2y}} = \frac{{3x}}{y}$

$\Rightarrow$ ${m_1} = \frac{{3x}}{y}$

and slope of normal $\left( {{m_2}} \right) = \frac{{ - 1}}{{{m_1}}} = \frac{{ - y}}{{3x}}$ ….(ii)

Since, slope of normal to the curve should be equal to the slope of line $x + 3y = 4$, which is parallel to curve.

For line, $y = \frac{{4 - x}}{3} = \frac{{ - x}}{3} + \frac{4}{3}$

$\Rightarrow$ Slope of the line $\left( {{m_3}} \right) = \frac{{ - 1}}{3}$

Therefore, ${m_2} = {m_3}$
$\Rightarrow$ $\frac{{ - y}}{{3x}} = - \frac{1}{3}$
$\Rightarrow$ $- 3y = - 3x$
$\Rightarrow$ $y = x$ …..(iii)

Now substituting the value of $y$ in Eq. (i), we get
$3{x^2} - {x^2} = 8$
$\Rightarrow$ ${x^2} = 4$
$\Rightarrow$ $x = \pm 2$
For $x = 2,$ $y = 2$ [using Eq. (iii)]

and for $x = - 2,$ $y = - 2$ [using Eq. (iii)]

Therefore the points at which normal to the curve are parallel to the line $x + 3y = 4$ are (2,2) and (-2,-2).

Hence the required equations of normal are
$y - 2 = {m_2}(x - 2)$ and $y + 2 = {m_2}(x + 2)$

$\Rightarrow$ $y - 2 = \frac{{ - 2}}{6}(x - 2)$ and $y + 2 = \frac{{ - 2}}{6}(x + 2)$

$\Rightarrow$ $3y - 6 = - x + 2\quad$ and $3y + 6 = - x - 2$

$\Rightarrow$ $3y + x = + 8$ and $3y + x = - 8$

So, the required equations are $3y + x = \pm 8$.