logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Prove that the curves \(x = y^2\) and \(xy = k\) cut at right angles* if \(8k^2 = 1.\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • 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}$$(x-x_1)$
Step 1:
The given curves :
$x=y^2$----(1)
$xy=k$-----(2)
Put $x=y^2$ in equ(2) we get,
$y^3=k$
$k^{\Large\frac{1}{3}}=y$
Step 2:
From equ(1) $x=y^2=\big(k^{\Large\frac{1}{3}}\big)^2$
$\Rightarrow x=k^{\Large\frac{2}{3}}$
Hence the point of intersection of the curves is $(k^{\Large\frac{1}{3}},k^{\Large\frac{2}{3}})$
Differentiate equ(1) w.r.t $x$ we get,
$1=2y\large\frac{dy}{dx}$
$\large\frac{dy}{dx}=\frac{1}{2y}$
Slope of the tangent at $(k^{\Large\frac{2}{3}},k^{\Large\frac{1}{3}})$
$\large\frac{dy}{dx}_{(k^{\Large\frac{2}{3}},k^{\Large\frac{1}{3}})}=\frac{1}{2k^{\Large\frac{1}{3}}}$
Step 3:
Differentiate equ(2) w.r.t $x$ we get,
$xy=k$
By applying product rule,
$x\large\frac{dy}{dx}+$$y.1=0$
$\Rightarrow \large\frac{dy}{dx}=-\large\frac{y}{x}$
Hence $\large\frac{dy}{dx}$ at $(k^{\Large\frac{2}{3}},k^{\Large\frac{1}{3}})$ is
$\large\frac{dy}{dx}_{(k^{\Large\frac{2}{3}},k^{\Large\frac{1}{3}})}=\frac{-k^{\Large\frac{1}{3}}}{k^{\Large\frac{2}{3}}}$
$\Rightarrow \large\frac{-1}{k^{\Large\frac{1}{3}}}$
Step 4:
If the curves cut at right angles,then the product of the slopes is $-1$
(i.e)$\large\frac{1}{2k^{\Large\frac{1}{3}}}.\frac{-1}{k^{\Large\frac{1}{3}}}$$=-1$
$\Rightarrow 1=2k^{\Large\frac{2}{3}}$
Cube on both sides,
$8k^2=1$
Hence proved.
answered Jul 12, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...