Question

The angle of intersection of the curves xy = a² and x² − y² = 2a² is

(a) 0°
(b) 45°
(c) 90°
(d) none of these

Answer 1

(c) 90°

$$\begin{gathered} \mathrm{Given}: \\ xy=a^2...\left(1\right) \\ {x}^2-y^2=2a^2...\left(2\right) \\ \text{Let}\left(x_1,y_1\right)\text{be the point of intersection.} \\ \mathrm{On}\mathrm{d}\text{ifferentiating (1) w.r.t.}\mathit{\text{x}}\text{, we get} \\ x\frac{dy}{dx}+y\mathit{=}\mathit{0} \\ \Rightarrow \frac{dy}{dx}=\frac{\mathit{-}y}{x} \\ \Rightarrow m_{\mathit{1}}={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{\mathit{-}{y}_1}{x_1} \\ \mathrm{On}\mathrm{d}\text{ifferentiating (2) w.r.t.}\mathit{\text{x}}\text{, we get} \\ 2x\mathit{}\mathit{-}2y\mathit{}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{x}{y} \\ \mathit{\Rightarrow }{m}_{\mathit{2}}={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=\frac{x_1}{y_1} \\ \mathrm{Now}, \\ {m}_1\times m_2=\frac{\mathit{-}{y}_{\mathit{1}}}{x_{\mathit{1}}}\times \frac{x_1}{y_1} \\ \Rightarrow m_1\times m_2=-1 \\ \because m_1\times m=-1 \\ \text{So, the angle between the curves is 90°.} \end{gathered}$$