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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The curves $y=4x^2+2x-8$ and $y=x^3-x+13$ touch each other at the point_________.

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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
Let curve $c_1$ be $ y=4x^2+2x-8$
Let curve $c_2$ be $ y=x^3-x+13$
Differentiating curve $c_1$ w.r.t $x$ we get,
$ \large\frac{dy}{dx}=8x+2$
Differentiating curve $c_2$ w.r.t $x$ we get,
$ \large\frac{dy}{dx}=3x^2-1$
Since they touch each other, let us assume that the slope of the tangent are equal at $(x_1, y_1)$
i.e., $8x_1+2 = 3x_1^2-1$
$ \Rightarrow 3x_1^2-8x_1-3=0$
On factorizing we get,
$(3x_1+1)(x-3), \: \Rightarrow x_1=-\large\frac{1}{3} \: and \: x=3$
Since $x_1 \neq - \large\frac{1}{3}$, let us consider $ x = 3$
$ \therefore y_1 = 4(3)^2+2(3)-8$
$ = 34$
Hence the point of intersection is $(3, 34)$
answered Aug 13, 2013 by thanvigandhi_1
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