Evaluate the distance between the points `(asinalpha,-bcosalpha)and(-acosalpha,bsinalpha)`.

Options

A. 2
B. `sqrt(a^(2)+b^(2))`
C. `2sqrt(a^(2)+b^(2))`
D. `sqrt(a^(2)+b^(2))(sinalpha+cosalpha)`

Answer

To find the distance between the points \((a \sin \alpha, -b \cos \alpha)\) and \((-a \cos \alpha, b \sin \alpha)\), we will use the distance formula. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

### Step-by-Step Solution:

1. **Identify the Coordinates**:
Let \(P = (a \sin \alpha, -b \cos \alpha)\) and \(Q = (-a \cos \alpha, b \sin \alpha)\).
Here, \(x_1 = a \sin \alpha\), \(y_1 = -b \cos \alpha\), \(x_2 = -a \cos \alpha\), and \(y_2 = b \sin \alpha\).

**Hint**: Write down the coordinates clearly to avoid confusion.

2. **Substitute into the Distance Formula**:
Substitute the coordinates into the distance formula:
\[ d = \sqrt{((-a \cos \alpha) - (a \sin \alpha))^2 + (b \sin \alpha - (-b \cos \alpha))^2} \]

**Hint**: Make sure to carefully handle the signs when substituting the coordinates.

3. **Simplify the Expressions**:
Calculate \(x_2 - x_1\) and \(y_2 - y_1\):
\[ x_2 - x_1 = -a \cos \alpha - a \sin \alpha = -a (\cos \alpha + \sin \alpha) \]
\[ y_2 - y_1 = b \sin \alpha + b \cos \alpha = b (\sin \alpha + \cos \alpha) \]

**Hint**: Factor out common terms to simplify the expressions.

4. **Square the Differences**:
Now, square both differences:
\[ (x_2 - x_1)^2 = (-a (\cos \alpha + \sin \alpha))^2 = a^2 (\cos \alpha + \sin \alpha)^2 \]
\[ (y_2 - y_1)^2 = (b (\sin \alpha + \cos \alpha))^2 = b^2 (\sin \alpha + \cos \alpha)^2 \]

**Hint**: Use the identity \((a + b)^2 = a^2 + 2ab + b^2\) to expand the squares.

5. **Combine the Squares**:
Combine the two squared terms:
\[ d^2 = a^2 (\cos \alpha + \sin \alpha)^2 + b^2 (\sin \alpha + \cos \alpha)^2 \]
Factor out \((\sin \alpha + \cos \alpha)^2\):
\[ d^2 = (a^2 + b^2)(\sin \alpha + \cos \alpha)^2 \]

**Hint**: Look for common factors to simplify the expression.

6. **Take the Square Root**:
Finally, take the square root to find the distance:
\[ d = \sqrt{(a^2 + b^2)(\sin \alpha + \cos \alpha)^2} = \sqrt{a^2 + b^2} \cdot |\sin \alpha + \cos \alpha| \]

**Hint**: Remember that the square root of a square gives the absolute value.

### Final Answer:
The distance between the points \((a \sin \alpha, -b \cos \alpha)\) and \((-a \cos \alpha, b \sin \alpha)\) is:
\[ d = \sqrt{a^2 + b^2} \cdot |\sin \alpha + \cos \alpha| \]