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

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

$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.