Reduce the following equations to the normal form and find the values of p and `alpha`.
`sqrt(3)x-y+2=0`

Answer

To reduce the equation \( \sqrt{3}x - y + 2 = 0 \) to its normal form and find the values of \( p \) and \( \alpha \), we can follow these steps:

### Step 1: Rewrite the equation
Start with the given equation:
\[ \sqrt{3}x - y + 2 = 0 \]
We can rearrange it to isolate \( y \):
\[ \sqrt{3}x - y = -2 \quad \Rightarrow \quad -\sqrt{3}x + y = -2 \]

### Step 2: Normalize the equation
Next, we want to express the equation in the normal form, which is:
\[ x \cos \alpha + y \sin \alpha = p \]
To do this, we need to divide the entire equation by the magnitude of the coefficients of \( x \) and \( y \). The coefficients are \( -\sqrt{3} \) for \( x \) and \( 1 \) for \( y \).

Calculate the magnitude:
\[ \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \]

Now, divide the entire equation by \( 2 \):
\[ \frac{-\sqrt{3}}{2}x + \frac{1}{2}y = -\frac{2}{2} \quad \Rightarrow \quad -\frac{\sqrt{3}}{2}x + \frac{1}{2}y = -1 \]

### Step 3: Compare with the normal form
Now we can compare this with the normal form:
\[ x \cos \alpha + y \sin \alpha = p \]
From our equation, we have:
\[ -\frac{\sqrt{3}}{2} = \cos \alpha \quad \text{and} \quad \frac{1}{2} = \sin \alpha \]
This gives us the values for \( \cos \alpha \) and \( \sin \alpha \).

### Step 4: Calculate \( \tan \alpha \)
To find \( \tan \alpha \):
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \]

### Step 5: Determine \( \alpha \)
The value \( \tan \alpha = -\frac{1}{\sqrt{3}} \) corresponds to angles in the second quadrant. The reference angle is \( 30^\circ \), so:
\[ \alpha = 180^\circ - 30^\circ = 150^\circ \]
In radians, this is:
\[ \alpha = \frac{5\pi}{6} \]

### Step 6: Find \( p \)
From our normalized equation, we see that:
\[ p = -1 \]

### Final Results
Thus, the values are:
\[ p = -1, \quad \alpha = 150^\circ \text{ or } \frac{5\pi}{6} \]