If the curves y = 2 ex and y = ae−x intersect orthogonally, then a = _____________ .

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$\frac{1}{2}$ \text { Given }: y = 2 e^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 { On differentiating (1) w.r.t.x, we get } dy/dx = 2 e^x \Rightarrow m_1 = \left( dy/dx \right)_\left( x_1 , y_1 \right) = 2 e^{x_1} \text { Now, on differentiating (2) w.r.t.x, we get } dy/dx = - a e^{- x} \Rightarrow m_2 = \left( 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 2 e^{x_1} $\times$ \left( - a e^{- x_1} \right) = - 1 \Rightarrow 2a = 1 \Rightarrow a = $\frac{1}{2}$

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