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The value of $'a'$ so that the curves $y=3e^{x}$ and $y=\large\frac{a}{3}e^{x}$ intersect orthogonally is

\[\begin{array}{1 1}(1)-1&(2)1\\(3)\frac{1}{3}&(4)3\end{array}\]

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$y=3e^{x}=> y=\large\frac{a}{3}$$e^{-x}$
$\large\frac{dy}{dx} $$=3e^x$
$\large\frac{dy}{dx}=\frac{a}{3} $$ \times e^{-x} \times -1$
$m_1=3e^x$
$m_2=-\large\frac{a}{3} e^{-x}$
$m_1m_2=-1$
$3e^x \times \large\frac{-a}{3} $$e^{-x}=-1$
$-a=-1$
$a=1$
Hence 2 is the correct answer.
answered May 19, 2014 by meena.p
 

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