The coordinates of the point at which the line `x cos alpha + y sin alpha + a sin alpha = 0` touches the parabola `y^(2) = 4x` are

Options

A. `(a tan alpha, 2a tan alpha)`
B. `(2a tan alpha, a tan alpha)`
C. `(a tan alpha, -2a )`
D. `(a cos alpha, 2a cos alpha)`

Answer

To find the coordinates of the point at which the line \( x \cos \alpha + y \sin \alpha + a \sin \alpha = 0 \) touches the parabola \( y^2 = 4x \), we can follow these steps:

### Step 1: Understand the Parametric Form of the Parabola
The parabola \( y^2 = 4x \) can be represented in parametric form as:
\[ P(t) = (bt^2, 2bt) \]
where \( b = 1 \) for the parabola \( y^2 = 4x \).

### Step 2: Write the Equation of the Tangent to the Parabola
The equation of the tangent to the parabola at the point \( P(t) \) is given by:
\[ ty = x + bt^2 \]
Rearranging gives:
\[ x - ty + bt^2 = 0 \]

### Step 3: Compare with the Given Line Equation
The given line equation is:
\[ x \cos \alpha + y \sin \alpha + a \sin \alpha = 0 \]
Rearranging gives:
\[ x \cos \alpha + y \sin \alpha = -a \sin \alpha \]
We can rewrite it as:
\[ x + \left( \frac{y \sin \alpha}{\cos \alpha} \right) = -\frac{a \sin \alpha}{\cos \alpha} \]
This can be compared with the tangent equation \( x - ty + bt^2 = 0 \).

### Step 4: Equate Coefficients
From the tangent equation \( x - ty + bt^2 = 0 \) and the line equation, we can equate coefficients:
1. Coefficient of \( x \): \( 1 = \cos \alpha \)
2. Coefficient of \( y \): \( -t = \sin \alpha \)
3. Constant term: \( bt^2 = -a \sin \alpha \)

### Step 5: Solve for \( t \)
From the second equation:
\[ t = -\frac{\sin \alpha}{\cos \alpha} = -\tan \alpha \]

### Step 6: Substitute \( t \) into the Constant Term Equation
Substituting \( t \) into the third equation:
\[ b(-\tan \alpha)^2 = -a \sin \alpha \]
Since \( b = 1 \):
\[ -\tan^2 \alpha = -a \sin \alpha \]
This simplifies to:
\[ \tan^2 \alpha = a \sin \alpha \]

### Step 7: Find the Coordinates of the Point of Tangency
Now we can find the coordinates of point \( P \):
\[ P(t) = (bt^2, 2bt) = (1 \cdot (-\tan \alpha)^2, 2 \cdot 1 \cdot (-\tan \alpha)) \]
Thus:
\[ P(t) = (\tan^2 \alpha, -2\tan \alpha) \]

### Step 8: Substitute \( b \) and Simplify
Using \( b = \frac{a}{\tan \alpha} \):
\[ P(t) = \left( \frac{a}{\tan \alpha}, -2a \right) \]

### Final Coordinates
The coordinates of the point at which the line touches the parabola are:
\[ \left( a \tan \alpha, -2a \right) \]