Question

Show that the curves 2x = y² and 2xy = k cut at right angles, if k² = 8.

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ 2x=y^2...\left(1\right) \\ 2xy=k...\left(2\right) \\ \text{From (1) and (2), we get} \\ {y}^3=k \\ \Rightarrow y=k^{\frac{1}{3}} \\ \text{From (1), we get} \\ 2x=k^{\frac{2}{3}} \\ \Rightarrow x=\frac{k^{\frac{2}{3}}}{2} \\ \text{On differentiating (1) w.r.t.}\mathit{\text{x,}}\text{we get} \\ 2=2y\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{y} \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(\frac{k^{\frac{2}{3}}}{2},k^{\frac{1}{3}}\right)}=\frac{1}{k^{\frac{1}{3}}}=k^{\frac{-1}{3}} \\ \text{On differentiating (2) w.r.t}\mathit{\text{.}}\mathit{\text{x,}}\text{we get} \\ 2x\frac{dy}{dx}+2y=0 \\ \Rightarrow \frac{dy}{dx}=\frac{\mathit{-}y}{x} \\ \mathit{\Rightarrow }{m}_2={\left(\frac{dy}{dx}\right)}_{\left(\frac{k^{\frac{2}{3}}}{2},k^{\frac{1}{3}}\right)}=\frac{-k^{\frac{1}{3}}}{\left(\frac{k^{\frac{2}{3}}}{2}\right)}=\mathit{-}\mathit{2}{k}^{\frac{-1}{3}} \\ \text{It is given that the curves intersect orthogonally}\mathit{\text{.}} \\ \therefore m_1\times m_2=-1 \\ \Rightarrow k^{\frac{-1}{3}}\times -2k^{\frac{-1}{3}}=-1 \\ \Rightarrow 2k^{\frac{-2}{3}}=1 \\ \Rightarrow k^{\frac{-2}{3}}=\frac{1}{2} \\ \Rightarrow k^{\frac{2}{3}}=2 \\ \text{Cubing on both sides, we get} \\ {k}^2=8 \end{gathered}$$