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# The equation of normal to the curve $3x^2-y^2=8$ which is parallel to the line x+3y=8 is\begin{array}{1 1}(A)\;3x-y=8 & (B)\;3x+y+8=0\\(C)\;x+3y-8=0 & (D)\;x+3y=0\end{array}

Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the normal to $y = f(x)$ at point $p$
Step 1
The equation of the curve is
$3x^2-y^2=8 \: \: \: (1)$
Differentiating w.r.t $x$ we get,
$6x-2y\large\frac{dy}{dx}=0$
$\Rightarrow \large\frac{dy}{dx}=\large\frac{3x}{y}$
It is given that the normal is parallel to the line.
$x +3y=8$
Slope of the line =
$-\large\frac{Coefficient \: of \: x}{Coefficient\: of \: y}$
$m=-\large\frac{1}{3}$
$\therefore$ slope of the normal is 3
$( \because m_1m_2=-1)$
Step 2
$\Rightarrow \large\frac{3x}{y}=3$
$\Rightarrow x =y$
$\therefore x = y$
Substituting this in equation (1) we get,
$3y^2-y^2=8$
$2y^2=8$
$y^2=4$
$y = \pm 4$
$\therefore$ The required points are $(4,4), (-4, -4)$
Step 3
Hence the equation of the normal at $(4,4)$
$(y-4)=3(x-4)$
$\Rightarrow 3x-y=8$
hence the correct option is A