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Solution

For $y = ae^{-x}$, $\dfrac{dy}{dx} = -ae^{-x} = -y$. For $y = be^{x}$, $\dfrac{dy}{dx} = be^{x} = y$. At point of intersection: $ae^{-x} = be^{x} \Rightarrow e^{2x} = \dfrac{a}{b}$. Slopes: $m_1 = -\dfrac{b}{a}$ and $m_2 = \dfrac{a}{b}$. For orthogonality: $m_1 m_2 = -1 \Rightarrow ab = 1$.