The length of the normal to the curve `y=a((e^(-x//a)+e^(x//a))/(2))` at any point varies as the

Options

A. Abscissa of the point
B. Ordinate of the point
C. Square of the abscissa of the point
D. Square of the ordinate of the point

Answer

To find the length of the normal to the curve \( y = a \left( \frac{e^{-\frac{x}{a}} + e^{\frac{x}{a}}}{2} \right) \) at any point, we will follow these steps:

### Step 1: Differentiate the function to find \( \frac{dy}{dx} \)

Given the function:
\[ y = a \left( \frac{e^{-\frac{x}{a}} + e^{\frac{x}{a}}}{2} \right) \]

We can differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = a \cdot \frac{1}{2} \left( \frac{d}{dx}(e^{-\frac{x}{a}}) + \frac{d}{dx}(e^{\frac{x}{a}}) \right) \]
Using the chain rule:
\[ \frac{d}{dx}(e^{-\frac{x}{a}}) = -\frac{1}{a} e^{-\frac{x}{a}}, \quad \frac{d}{dx}(e^{\frac{x}{a}}) = \frac{1}{a} e^{\frac{x}{a}} \]
Thus,
\[ \frac{dy}{dx} = a \cdot \frac{1}{2} \left( -\frac{1}{a} e^{-\frac{x}{a}} + \frac{1}{a} e^{\frac{x}{a}} \right) = \frac{1}{2} \left( e^{\frac{x}{a}} - e^{-\frac{x}{a}} \right) \]

### Step 2: Find the slope of the normal

The slope of the tangent line at any point is given by \( \frac{dy}{dx} \). Therefore, the slope of the normal line is:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{2}{e^{\frac{x}{a}} - e^{-\frac{x}{a}}} \]

### Step 3: Length of the normal

The length \( L \) of the normal at any point can be expressed as:
\[ L = y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \]
Substituting \( y \) and \( \frac{dy}{dx} \):
\[ L = a \left( \frac{e^{-\frac{x}{a}} + e^{\frac{x}{a}}}{2} \right) \sqrt{1 + \left( \frac{1}{2} \left( e^{\frac{x}{a}} - e^{-\frac{x}{a}} \right) \right)^2} \]
Calculating \( 1 + \left( \frac{1}{2} \left( e^{\frac{x}{a}} - e^{-\frac{x}{a}} \right) \right)^2 \):
\[ = 1 + \frac{1}{4} \left( e^{\frac{x}{a}} - e^{-\frac{x}{a}} \right)^2 \]
This simplifies to:
\[ = 1 + \frac{1}{4} \left( e^{\frac{2x}{a}} - 2 + e^{-\frac{2x}{a}} \right) = \frac{4 + e^{\frac{2x}{a}} - 2 + e^{-\frac{2x}{a}}}{4} = \frac{2 + e^{\frac{2x}{a}} + e^{-\frac{2x}{a}}}{4} \]
Thus:
\[ L = a \left( \frac{e^{-\frac{x}{a}} + e^{\frac{x}{a}}}{2} \right) \sqrt{\frac{2 + e^{\frac{2x}{a}} + e^{-\frac{2x}{a}}}{4}} = a \left( \frac{e^{-\frac{x}{a}} + e^{\frac{x}{a}}}{2} \right) \cdot \frac{\sqrt{2 + 2 \cosh\left(\frac{2x}{a}\right)}}{2} \]

### Step 4: Final expression for the length of the normal

After simplifying, we find that:
\[ L = \frac{a}{2} \left( e^{-\frac{x}{a}} + e^{\frac{x}{a}} \right) \sqrt{2 + 2 \cosh\left(\frac{2x}{a}\right)} \]

### Conclusion

The length of the normal to the curve at any point varies as \( y^2 \).