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

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
$3x^2-y^2=8\: \: \: (1)$
The given line is $x+3y=4$
consider the equation
$3x^2-y^2=8$
differentiate w.r.t $x$ we get
$6x-2y\large\frac{dy}{dx}=0$
$\Rightarrow 2y\large\frac{dy}{dx}=6x$
$\large\frac{dy}{dx}=\large\frac{3x}{y}$
This is the slope of the tangent to the given curve.
$\therefore$ slope of the normal is $-\large\frac{y}{3x}$
The slope of the given line is $\bigg( \large\frac{-coefficient\: of\: x}{coefficient\: of\: y} \bigg)$
i.e., $-\large\frac{1}{3}$
Step 2
Since the normal is parallel to the given line, the slopes are equal
$\therefore -\large\frac{y}{3x}=-\large\frac{1}{3}$
$\Rightarrow y=x$
Substituting for $y$ in equation (1) we get
$3x^2-x^2=8$
$2x^2=8$
$x^2=4$
$\therefore x= \pm 2$
$\Rightarrow y= \pm 2$
hence the points of intersection are $(\pm2, \pm2)$
Step 3
Hence the equation of the normal is when $(x,y) = (2,2)$
$(y-y_1)=-\large\frac{1}{m}(x-x_1)$
$\Rightarrow (y-2)=-\large\frac{1}{3}(x-2)$
$3y-6=-x+2$
$\Rightarrow x+3y-8=0$
Step 4
When $(x,y)=(-2,-2)$
$(y+2)=-\large\frac{1}{3}(x+2)$
$\Rightarrow x+3y+8=0$
Hence the equations of the normal are
$x+3y \pm8=0$