Question

The value of '$a$' so that the curves $y=3e^x$ and $y=\frac{a}{3}{e}^{-x}$ are perpendicular to each other:

• A. $\frac{1}{3}$
• B. $-1$
• C. $3$
• D. $1$

Answer 1

The correct option is D $1$

Given:$y=3e^x$

$\Rightarrow m_1=\frac{dy}{dx}=3e^x$

Given:$y=\frac{a}{3}{e}^{-x}$

$\Rightarrow m_2=\frac{dy}{dx}=\frac{-a}{3}{e}^{-x}$

The curves intersect orthogonally if the product pf their slopes$-1$

$\Rightarrow m_1{m}_2=-1$

$\Rightarrow 3e^x\times \frac{-a}{3}{e}^{-x}=-1$

$\Rightarrow a=1$

$\therefore a=1$