Which of the following pair (s) `is//are` orthogonal ?

Options

A. `16x^(2)+y^(2)=c and y^(16)=kx`
B. `y=x+ce^(-x)and x+2=y+ke^(-y)`
C. `y=Cx^(2)and x^(2)+2y^(2)=k`
D. `x^(2)-y^(2)=C and xy=k`

Answer

To determine which of the given pairs of functions are orthogonal, we need to follow the definition of orthogonality for functions. Two functions \( f(x) \) and \( g(x) \) are said to be orthogonal if the derivatives \( \frac{dy}{dx} \) of \( f(x) \) and \( \frac{dx}{dy} \) of \( g(x) \) are equal.

Let's analyze each option step by step.

### Option 1:
Given functions:
1. \( 16x^2 + y^2 = c \)
2. \( y^{16} = kx \)

**Step 1: Differentiate the first function with respect to \( x \)**

\[ \frac{d}{dx}(16x^2 + y^2) = 0 \implies 32x + 2y\frac{dy}{dx} = 0 \]

Rearranging gives:

\[ 2y\frac{dy}{dx} = -32x \implies \frac{dy}{dx} = -\frac{16x}{y} \]

**Step 2: Differentiate the second function with respect to \( y \)**

\[ \frac{d}{dy}(y^{16}) = k \implies 16y^{15} = k\frac{dx}{dy} \]

Rearranging gives:

\[ \frac{dx}{dy} = \frac{16y^{15}}{k} \]

**Step 3: Check for equality**

We need to check if:

\[ -\frac{16x}{y} = \frac{16y^{15}}{k} \]

This equality does not hold for all \( x \) and \( y \), so the functions are **not orthogonal**.

### Option 2:
Given functions:
1. \( y = x + ce^{-x} \)
2. \( x + 2 = y + ke^{-y} \)

**Step 1: Differentiate the first function with respect to \( x \)**

\[ \frac{dy}{dx} = 1 - ce^{-x} \]

**Step 2: Differentiate the second function with respect to \( y \)**

\[ \frac{d}{dy}(x + 2) = \frac{d}{dy}(y + ke^{-y}) \implies 0 = 1 - ke^{-y}\frac{dx}{dy} \]

Rearranging gives:

\[ \frac{dx}{dy} = \frac{1}{ke^{-y}} = \frac{e^{y}}{k} \]

**Step 3: Check for equality**

To be orthogonal, we need:

\[ 1 - ce^{-x} = \frac{e^{y}}{k} \]

This holds true if \( c = k \). Thus, the functions can be orthogonal under this condition.

### Option 3:
Given functions:
1. \( y = cx^2 \)
2. \( x^2 + 2y^2 = k \)

**Step 1: Differentiate the first function with respect to \( x \)**

\[ \frac{dy}{dx} = 2cx \]

**Step 2: Differentiate the second function with respect to \( y \)**

\[ \frac{d}{dy}(x^2 + 2y^2) = 0 \implies 2y\frac{dy}{dx} + 2x\frac{dx}{dy} = 0 \]

Rearranging gives:

\[ \frac{dx}{dy} = -\frac{y}{x} \]

**Step 3: Check for equality**

We need to check if:

\[ 2cx = -\frac{y}{x} \]

This equality does not hold for all \( x \) and \( y \), so the functions are **not orthogonal**.

### Option 4:
Given functions:
1. \( x^2 - y^2 = C \)
2. \( xy = k \)

**Step 1: Differentiate the first function with respect to \( x \)**

\[ \frac{d}{dx}(x^2 - y^2) = 0 \implies 2x - 2y\frac{dy}{dx} = 0 \]

Rearranging gives:

\[ \frac{dy}{dx} = \frac{x}{y} \]

**Step 2: Differentiate the second function with respect to \( y \)**

\[ \frac{d}{dy}(xy) = 0 \implies x + y\frac{dx}{dy} = 0 \]

Rearranging gives:

\[ \frac{dx}{dy} = -\frac{x}{y} \]

**Step 3: Check for equality**

We need to check if:

\[ \frac{x}{y} = -\frac{x}{y} \]

This equality does not hold, so the functions are **not orthogonal**.

### Conclusion:
After analyzing all options, we find that only **Option 2** can be orthogonal under the condition \( c = k \). The other options do not yield orthogonal pairs.