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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equation of the normal lines to the curve $3x^2-y^2=8$ which are parallel to the line x+3y=4.

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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$
answered Aug 4, 2013 by thanvigandhi_1
 

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