Question

The angle between the curves y² = x and x² = y at (1, 1) is

(a) ${\tan }^{-1}\frac{4}{3}$
(b) ${\tan }^{-1}\frac{3}{4}$
(c) 90°
(d) 45°

Answer 1

(b) ${\tan }^{-1}\frac{3}{4}$

$$\begin{gathered} \mathrm{Given}: \\ {y}^2=x...\left(1\right) \\ {x}^2=y...\left(2\right) \\ \mathrm{Point}=\left(1,1\right) \\ \mathrm{On}\mathrm{d}\text{ifferentiating (1) w.r.t.}\mathit{\text{x}}\text{, we get} \\ \mathit{2}y\mathit{}\frac{dy}{dx}=1 \\ \mathit{\Rightarrow }\frac{dy}{dx}=\frac{\mathit{1}}{\mathit{2}y} \\ \mathit{\Rightarrow }{m}_1=\frac{1}{2} \\ \mathrm{On}\mathrm{d}\text{ifferentiating (2) w.r.t.}\mathit{\text{x}}\text{, we get} \\ 2x\mathit{}=\mathit{}\frac{dy}{dx} \\ \mathit{\Rightarrow }{m}_{\mathit{2}}=2\left(1\right)\mathit{=}2 \\ \mathrm{Now}, \\ \tan \mathit{}\theta \mathit{=}\left|\frac{m_{\mathit{1}}\mathit{-}{m}_{\mathit{2}}}{1+m_{\mathit{1}}{m}_{\mathit{2}}}\right|\mathit{=}\left|\frac{\frac{1}{2}\mathit{-}\mathit{2}}{1+\frac{1}{2}\times 2}\right|\mathit{=}\frac{3}{4} \\ \Rightarrow \theta \mathit{=}{\tan }^{-1}\mathit{}\left(\frac{3}{4}\right) \end{gathered}$$