#### The equation of normal to the curve $3x\textsuperscript{2} - y^2 = 8$ which is parallel to the line $x + 3y = 8$ is $\\A. 3x - y = 8\\ B. 3x + y + 8 = 0\\ C. x + 3y \pm 8 = 0 \\D. x + 3y = 0$

Given the equation of the line is  $3x\textsuperscript{2} - y^2 = 8$

Differentiate both sides with x and get

$\frac{\mathrm{d}\left(3 \mathrm{x}^{2}-\mathrm{y}^{2}\right)}{\mathrm{dx}}=\frac{\mathrm{d}(8)}{\mathrm{dx}}$
Apply sum rule and 0 is the differentiation of constant, so
$\Rightarrow \frac{\mathrm{d}\left(3 \mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0$
Take the constants out and get
$\Rightarrow 3 \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(\mathrm{y}^{2}\right)}{\mathrm{dx}}=0$
Apply power rule and get

$\Rightarrow 3 \times 2\left(x^{2-1}\right) \times \frac{d(x)}{dx}-2\left(y^{2-1}\right) \times \frac{d(y)}{dx}=0 \\ \Rightarrow 6x-2y \times \frac{dy}{dx}=0 \\ \Rightarrow 6x=2y \times \frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{6x}{2y}=\frac{3x}{y}$

Hence, the slope of the given curve is provided.

Also, the slope of the normal to the curve is

$=-\frac{1}{\frac{d y}{d x}} \\\Rightarrow=-\frac{1}{\frac{3 \mathrm{x}}{\mathrm{y}}}=\left(-\frac{\mathrm{y}}{3 \mathrm{x}}\right) \ldots \ldots (i)$
Now, $x+3 y=8$
$\Rightarrow 3 y=8-x$
After differentiating with respect to x
$\\\Rightarrow \frac{\mathrm{d}(3 \mathrm{y})}{\mathrm{dx}}=\frac{\mathrm{d}(8-\mathrm{x})}{\mathrm{dx}} \\\Rightarrow 3 \frac{\mathrm{dy}}{\mathrm{dx}}=-1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{3}$
Therefore, the slope is $-\frac{1}{3}$
Now, because the normal to the curve is parallel to this line, that means the slope of the line must be equal to slope of the normal to the given curve,
$\therefore\left(-\frac{y}{3 x}\right)=-\frac{1}{3}$

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

Substitute the value of the given equation

$\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3x\textsuperscript{2} - (x)\textsuperscript{2} = 8}\\ { \Rightarrow 2x\textsuperscript{2} = 8}\\ { \Rightarrow x\textsuperscript{2} = 4}\\ { \Rightarrow x= \pm 2}\\$

When x=2, the equation is

$\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(2)\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(4) - y\textsuperscript{2} = 8}\\ { \Rightarrow 12-8= y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4}\\ { \Rightarrow y= \pm 2}\\$

When x=-2, the equation is

$\\ {3x\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(-2)\textsuperscript{2} - y\textsuperscript{2} = 8}\\ { \Rightarrow 3(4) - y\textsuperscript{2} = 8}\\ { \Rightarrow 12-8= y\textsuperscript{2}}\\ { \Rightarrow y\textsuperscript{2} = 4}\\ { \Rightarrow y= \pm 2}\\$

So, the points are $( \pm 2, \pm 2)$ at which normal is parallel to the given line.

And required equation at $( \pm 2, \pm 2)$ is

$\\ y-(\pm 2)=-\frac{1}{3}[x-(\pm 2)] \\ \Rightarrow 3(y-(\pm 2))=-(x-(\pm 2)) \\ \Rightarrow 3 y-(\pm 6)=-x+(\pm 2) \\ \Rightarrow x+3 y-(\pm 6)-(\pm 2)=0 \\ \Rightarrow x+3 y+(\pm \quad 8)=0$

Hence the equation of normal to the curve is $x+3 y+(\pm \quad 8)=0$

So the correct answer is option C