The distance between a tangent to the parabola `y^2 = 4ax` which is inclined to axis at an angle `alpha` to X axis. and a parallel normal is

Options

A. `(acos alpha)/(sin^2alpha)`
B. `(a sin alpha)/(cos^2alpha)`
C. `a/(sin alpha cos^2alpha)`
D. `a/(cos alpha sin^2alpha)`

Answer

To solve the problem of finding the distance between a tangent to the parabola \( y^2 = 4ax \) inclined at an angle \( \alpha \) to the x-axis and a parallel normal, we will follow these steps:

### Step 1: Understand the equations of the tangent and normal
The general equation of a tangent to the parabola \( y^2 = 4ax \) at a point \( P(t_1) \) is given by:
\[ y = mx + \frac{a}{m} \]
where \( m = \tan(\alpha) \).

The equation of the normal at the same point is:
\[ y = mx - 2am - am^3 \]

### Step 2: Identify the equations of the lines
For the tangent line at point \( P(t_1) \):
\[ y = \tan(\alpha)x + \frac{a}{\tan(\alpha)} \]

For the normal line at point \( P(t_2) \) (which is parallel to the tangent):
\[ y = \tan(\alpha)x - 2a\tan(\alpha) - a\tan^3(\alpha) \]

### Step 3: Calculate the distance between the two parallel lines
The distance \( d \) between two parallel lines given by \( y = mx + c_1 \) and \( y = mx + c_2 \) is calculated using the formula:
\[ d = \frac{|c_2 - c_1|}{\sqrt{1 + m^2}} \]

Here, \( c_1 = \frac{a}{\tan(\alpha)} \) and \( c_2 = -2a\tan(\alpha) - a\tan^3(\alpha) \).

### Step 4: Substitute the values into the distance formula
Substituting \( c_1 \) and \( c_2 \) into the distance formula:
\[ d = \frac{\left| -2a\tan(\alpha) - a\tan^3(\alpha) - \frac{a}{\tan(\alpha)} \right|}{\sqrt{1 + \tan^2(\alpha)}} \]

### Step 5: Simplify the expression
First, simplify the numerator:
\[ = -2a\tan(\alpha) - a\tan^3(\alpha) - \frac{a}{\tan(\alpha)} = -a\left(2\tan(\alpha) + \tan^3(\alpha) + \frac{1}{\tan(\alpha)}\right) \]

The denominator simplifies using the identity \( 1 + \tan^2(\alpha) = \sec^2(\alpha) \):
\[ \sqrt{1 + \tan^2(\alpha)} = \sec(\alpha) \]

Thus, the distance \( d \) becomes:
\[ d = \frac{a\left| 2\tan(\alpha) + \tan^3(\alpha) + \frac{1}{\tan(\alpha)} \right|}{\sec(\alpha)} \]

### Step 6: Convert to sine and cosine
Using \( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} \) and \( \sec(\alpha) = \frac{1}{\cos(\alpha)} \):
\[ d = a\cos(\alpha)\left| 2\frac{\sin(\alpha)}{\cos(\alpha)} + \left(\frac{\sin^3(\alpha)}{\cos^3(\alpha)}\right) + \frac{\cos(\alpha)}{\sin(\alpha)} \right| \]

### Final Answer
After simplifying, we find:
\[ d = \frac{a}{\cos^2(\alpha) \sin(\alpha)} \]