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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the condition that the curves $2x=y^2$ and $2xy=k$ intersect orthogonally.

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  • When two lines are perpendicular then the product of the slopes is $-1$
  • If $y=f(x)$, then $\bigg( \large\frac{dy}{dx} \bigg)_p=$ slope of the tangent to $y=f(x)$ at a point $P.$
Step 1
The given curves are $ x=y^2 \: \: \: (1)\: and \: 2xy=k\: \: \: (2)$
Substitute for $x$ in the second equation.
$ 2y^3=k \Rightarrow y=\bigg(\large\frac{k}{2}\bigg)^{\large\frac{1}{3}}$
Put $ y=\bigg(\large\frac{k}{2}\bigg)^{\large\frac{1}{3}}, $ we get $x=\bigg(\large\frac{k}{2}\bigg)^{\large\frac{2}{3}}$
So the two curves intersect at the point.
Step 2
$ P\bigg( \bigg(\large\frac{k}{2}\bigg)^{\large\frac{2}{3}},\bigg(\large\frac{k}{2}\bigg)^{\large\frac{1}{3}} \bigg)$
Consider $x=y^2$
differentiating w.r.t $x$ we get
$1=2y.\large\frac{dy}{dr} \Rightarrow \large\frac{dy}{dx}=\large\frac{1}{2y}$
Let this be $m_1$
$ \therefore m_1=\large\frac{1}{\bigg(\large\frac{k}{2}\bigg)^{\large\frac{1}{3}}}=\large\frac{2^{\large\frac{2}{3}}}{k^{\large\frac{1}{3}}}$
Consider $2xy=k$
Differentiating w.r.t $x$ we get
$ 2y+2x\large\frac{dy}{dx}=0$
$ \Rightarrow \large\frac{dy}{dx}=-\large\frac{y}{x}=-\Large\frac{\bigg(\Large\frac{k}{2}\bigg)^{\Large\frac{1}{3}} }{\bigg(\Large\frac{k}{2}\bigg)^{\Large\frac{2}{3}} }=-\large\frac{2}{k^{\large\frac{1}{3}}}$
Let this be $m_2$
$ \therefore m_2=-\large\frac{2}{k^{\large\frac{1}{3}}}$
Since the curves cut at right angle, we have
$ \Rightarrow \large\frac{2^{\Large\frac{2}{3}}}{k^{\Large\frac{1}{3}}} \times -\Large\frac{2}{k^{\Large\frac{1}{3}}}=-1$
$ \Rightarrow \large\frac{k^{\Large\frac{2}{3}}}{2}=1$
cubing on both sides we get,
$ \bigg( \large\frac{k^{\large\frac{2}{3}}}{2} \bigg)^3=1^3$
$ \Rightarrow k^2=8$
hence the required condition is $k^2=8$
answered Aug 2, 2013 by thanvigandhi_1

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