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

Explanation

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

### Step 1: Find \( \frac{dy}{dx} \)

Given the curve:
\[ y = a \frac{e^{-x/a} + e^{x/a}}{2} \]

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

### Step 2: Find the Length of the Normal

The formula for the length of the normal at a point on the curve is given by:
\[ L = y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \]

First, we need to calculate \( \left( \frac{dy}{dx} \right)^2 \):
\[ \left( \frac{dy}{dx} \right)^2 = \left( \frac{1}{2} (e^{x/a} - e^{-x/a}) \right)^2 = \frac{1}{4} (e^{x/a} - e^{-x/a})^2 \]

Now, we can expand \( (e^{x/a} - e^{-x/a})^2 \):
\[ (e^{x/a} - e^{-x/a})^2 = e^{2x/a} - 2 + e^{-2x/a} \]
Thus,
\[ \left( \frac{dy}{dx} \right)^2 = \frac{1}{4} (e^{2x/a} - 2 + e^{-2x/a}) \]

Now, substituting this into the length of the normal formula:
\[ L = y \sqrt{1 + \frac{1}{4} (e^{2x/a} - 2 + e^{-2x/a})} \]

### Step 3: Substitute \( y \)

We know \( y = a \frac{e^{-x/a} + e^{x/a}}{2} \). Thus, we substitute \( y \) into the length of the normal:
\[ L = a \frac{e^{-x/a} + e^{x/a}}{2} \sqrt{1 + \frac{1}{4} (e^{2x/a} - 2 + e^{-2x/a})} \]

### Step 4: Simplify the Expression

Now, we simplify \( 1 + \frac{1}{4} (e^{2x/a} - 2 + e^{-2x/a}) \):
\[ 1 + \frac{1}{4} (e^{2x/a} - 2 + e^{-2x/a}) = \frac{4 + e^{2x/a} - 2 + e^{-2x/a}}{4} = \frac{2 + e^{2x/a} + e^{-2x/a}}{4} \]
Using the identity \( e^{x/a} + e^{-x/a} = 2 \cosh(x/a) \):
\[ L = a \frac{e^{-x/a} + e^{x/a}}{2} \sqrt{\frac{2 + 2 \cosh(2x/a)}{4}} = a \frac{e^{-x/a} + e^{x/a}}{2} \cdot \frac{\sqrt{2(1 + \cosh(2x/a))}}{2} \]

### Step 5: Final Expression

The length of the normal simplifies to:
\[ L = \frac{a}{2} (e^{-x/a} + e^{x/a}) \cdot \frac{\sqrt{2(1 + \cosh(2x/a))}}{2} \]

### Conclusion

The length of the normal to the curve varies as the square of the ordinate of the point, which is \( y^2 \).