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Question
If the curves y = 2 ex and y = ae−x intersect orthogonally, then a = _____________ .
Options
1/2
−1/2
2
2e2
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Solution
1/2
\[\text { Given }: \]
\[y = 2 e^x . . . \left( 1 \right)\]
\[y = a e^{- x} . . . \left( 2 \right)\]
\[\text { Let the point of intersection of these curves be }\left( x_1 , y_1 \right).\]
\[\text { On differentiating (1) w.r.t.x, we get }\]
\[\frac{dy}{dx} = 2 e^x \]
\[ \Rightarrow m_1 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = 2 e^{x_1} \]
\[\text { Now, on differentiating (2) w.r.t.x, we get }\]
\[\frac{dy}{dx} = - a e^{- x} \]
\[ \Rightarrow m_2 = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = - a e^{- x_1} \]
\[\text { It is given that the curves are orthogonal }.\]
\[ \therefore m_1 \times m_2 = - 1\]
\[ \Rightarrow 2 e^{x_1} \times \left( - a e^{- x_1} \right) = - 1\]
\[ \Rightarrow 2a = 1\]
\[ \Rightarrow a = \frac{1}{2}\]
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