Question

If the curves y = 2 e^x and y = ae^−x intersect orthogonally, then a =

(a) 1/2
(b) −1/2
(c) 2
(d) 2e²

Answer 1

(a) 1/2

$$\begin{gathered} \mathrm{Given}: \\ y=2e^x...\left(1\right) \\ y=a{e}^{-x}...\left(2\right) \\ \text{Let the point of intersection of these curves be}\left(x_1,y_1\right)\text{.} \\ \mathrm{On}\mathrm{d}\text{ifferentiating (1) w.r.t.}\mathit{\text{x}}\text{, we get} \\ \frac{dy}{dx}=2e^x \\ \Rightarrow m_1={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=2e^{x_1} \\ \mathrm{Now},\mathrm{on}\mathrm{d}\text{ifferentiating (2) w.r.t.}\mathit{\text{x}}\text{, we get} \\ \frac{dy}{dx}=-a{e}^{-x} \\ \Rightarrow m_2={\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=-a{e}^{-x_1} \\ \text{It is given that the curves are orthogonal.} \\ \therefore m_1\times m_2=-1 \\ \Rightarrow 2e^{x_1}\times \left(-a{e}^{-x_1}\right)=-1 \\ \Rightarrow 2a=1 \\ \Rightarrow a=\frac{1}{2} \end{gathered}$$