The value of 'a' so that the curves $y = 3e^x$ and $y = \dfrac{a}{3}e^{-x}$ intersect orthogonally is : (a) $\dfrac{1}{3}$ (b) $-1$ (c) $3$ (d) $1$

> [!TLDR] > Two curves intersect orthogonally when the product of their slopes at the common point is $-1$; computing this product for the given exponential curves gives $a=1$. > > 1. Differentiate $y_1 = 3e^x$: $y_1' = 3e^x$. > 2. Differentiate $y_2 = \dfrac{a}{3}e^{-x}$: $y_2' = -\dfrac{a}{3}e^{-x}$. > 3. Two curves cut orthogonally at a common point when the product of their slopes there equals $-1$: $y_1'\,y_2' = -1$. …

The value of 'a' so that the curves y = 3ex and y = (a)/(3)e-x intersect orthogonally is : (a) (1)/(3) (b) -1

Question 96 of 126

y = 3e^x

(a)

\dfrac{1}{3}

(b)

-1

(c)

3

(d)

1

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018MCQ· 1mImportance★★★★★

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Two curves intersect orthogonally when the product of their slopes at the common point is $-1$ ; computing this product for the given exponential curves gives $a=1$ .

Differentiate $y_1 = 3e^x$ : $y_1' = 3e^x$ .
Differentiate $y_2 = \dfrac{a}{3}e^{-x}$ : $y_2' = -\dfrac{a}{3}e^{-x}$ .
Two curves cut orthogonally at a common point when the product of their slopes there equals $-1$ : $y_1'\,y_2' = -1$ . …

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Q.The value of 'a' so that the curves $y = 3e^x$ and $y = \dfrac{a}{3}e^{-x}$ intersect orthogonally is :

Concept understanding — Orthogonal Trajectories

Orthogonal Trajectories

Q.The value of 'a' so that the curves y=3exy = 3e^xy=3ex and y=a3e−xy = \dfrac{a}{3}e^{-x}y=3a​e−x intersect orthogonally is :

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Q.The value of 'a' so that the curves $y = 3e^x$ and $y = \dfrac{a}{3}e^{-x}$ intersect orthogonally is :

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