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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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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}

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1 Answer

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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
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
answered Aug 9, 2013 by thanvigandhi_1
 

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