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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Two curves $x^3-3xy^2+2=0$ and $3x^2-y^3-2=0$ intersect at an angle of


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  • 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 given curves are $ x^3-3xy^2+2=0\: and \: 3x^2y-y^3-2=0$
Consider the equation
$ x^3-3xy^2+2=0$
differentiating w.r.t $x$ we get,
$ 3x^2-(3x.2y.\large\frac{dy}{dx}+y^2.3)+0=0$
$ \Rightarrow 3x^2-6xy.\large\frac{dy}{dx}-3y^2=0$
$ \therefore \large\frac{dy}{dx}=\large\frac{3y^2-3x^2}{-6xy}$
$ -\large\frac{(y^2-x^2)}{2xy}$
Let this be $m_1$
Step 2
Consider the equation
$ 3x^2-y^3-2=0$
differentiating w.r.t $x$ we get,
$ 3x^2.\large\frac{dy}{dx}+y.6x-3y^2.\large\frac{dy}{dx}=0$
$ \Rightarrow x^2.\large\frac{dy}{dx}+2xy-y^2.\large\frac{dy}{dx}=0$
$ \Rightarrow \large\frac{dy}{dx}(x^2-y^2)=-2xy$
$ \Rightarrow \large\frac{dy}{dx}=\large\frac{-2xy}{x^2-y^2}$
Let this be $m_2$
The product of the slopes is $m_1 \times m_2$
$ = -\bigg( \large\frac{y^2-x^2}{2xy} \bigg) \times \large\frac{-2xy}{x^2-y^2}$
$ = -1$
Since the product of the slopes is $-1$, the angle between them is $ \large\frac{\pi}{2}$
Hence C is the correct option.
answered Aug 13, 2013 by thanvigandhi_1
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