Question:
If the curves $y=2 e^{x}$ and $y=a e^{-x}$ interest orthogonally, then $a=$
A. $\frac{1}{2}$
B. $-\frac{1}{2}$
C. 2
D. $2 e^{2}$
Solution:
Given that the curves $y=2 e^{x}$ and $y=a e^{-x}$
Differentiating both of them w.r.t. $x$,
$\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{e}^{\mathrm{x}}$ and $\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{ae}^{-\mathrm{x}}$
Let $\mathrm{m}_{1}=2 \mathrm{e}^{\mathrm{x}}$ and $\mathrm{m}_{2}=-\mathrm{ae}^{-\mathrm{x}}$
$m_{1} \times m_{2}=-1$
(Because curves cut each other orthogonally)
$\Rightarrow-2 a=-1$
$\Rightarrow a=\frac{1}{2}$
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