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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Prove that the curves \(x = y^2\) and \(xy = k\) cut at right angles* if \(8k^2 = 1.\)

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  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
The given curves :
Put $x=y^2$ in equ(2) we get,
Step 2:
From equ(1) $x=y^2=\big(k^{\Large\frac{1}{3}}\big)^2$
$\Rightarrow x=k^{\Large\frac{2}{3}}$
Hence the point of intersection of the curves is $(k^{\Large\frac{1}{3}},k^{\Large\frac{2}{3}})$
Differentiate equ(1) w.r.t $x$ we get,
Slope of the tangent at $(k^{\Large\frac{2}{3}},k^{\Large\frac{1}{3}})$
Step 3:
Differentiate equ(2) w.r.t $x$ we get,
By applying product rule,
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{y}{x}$
Hence $\large\frac{dy}{dx}$ at $(k^{\Large\frac{2}{3}},k^{\Large\frac{1}{3}})$ is
$\Rightarrow \large\frac{-1}{k^{\Large\frac{1}{3}}}$
Step 4:
If the curves cut at right angles,then the product of the slopes is $-1$
$\Rightarrow 1=2k^{\Large\frac{2}{3}}$
Cube on both sides,
Hence proved.
answered Jul 12, 2013 by sreemathi.v

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