Browse Questions

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

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.