The locus of point of intersection of any tangent to the parabola `y ^(2) = 4a (x -2)` with a line perpendicular to it and passing through the focus, is

Options

A. The tangent to the parabola at the vertex
B. `x =2`
C. `x =0`
D. none of these

Answer

To find the locus of the point of intersection of any tangent to the parabola \( y^2 = 4a(x - 2) \) with a line perpendicular to it and passing through the focus, we can follow these steps:

### Step 1: Identify the Parabola and Its Properties
The given parabola is \( y^2 = 4a(x - 2) \). This is a standard form of a parabola that opens to the right. The vertex of the parabola is at the point \( (2, 0) \) and the focus is at \( (2 + a, 0) \).

**Hint:** Remember that the focus of a parabola in the form \( y^2 = 4px \) is located at \( (p, 0) \) from the vertex.

### Step 2: Write the Equation of the Tangent
The equation of the tangent to the parabola at a point \( (2 + at^2, 2at) \) can be derived using the formula for the tangent to the parabola:
\[ ty = 2a(x - 2 + at^2) \]
Rearranging gives:
\[ ty = 2ax - 4a + 2at^2 \]

**Hint:** Use the point-slope form of the tangent line to find the equation.

### Step 3: Find the Slope of the Tangent
The slope of the tangent line at the point \( (2 + at^2, 2at) \) is given by \( \frac{dy}{dx} = \frac{2a}{t} \). Therefore, the slope of the line perpendicular to this tangent will be \( -\frac{t}{2a} \).

**Hint:** The product of the slopes of two perpendicular lines is -1.

### Step 4: Write the Equation of the Perpendicular Line
The equation of the line perpendicular to the tangent and passing through the focus \( (2 + a, 0) \) can be written as:
\[ y - 0 = -\frac{t}{2a}(x - (2 + a)) \]
This simplifies to:
\[ y = -\frac{t}{2a}(x - (2 + a)) \]

**Hint:** Use the point-slope form of a line to write this equation.

### Step 5: Find the Intersection of the Tangent and the Perpendicular Line
To find the intersection of the tangent and the perpendicular line, we set the equations equal to each other:
\[ ty = 2ax - 4a + 2at^2 \]
and
\[ y = -\frac{t}{2a}(x - (2 + a)) \]
Substituting the second equation into the first gives:
\[ t\left(-\frac{t}{2a}(x - (2 + a))\right) = 2ax - 4a + 2at^2 \]

**Hint:** Substitute one equation into the other to find the coordinates of the intersection.

### Step 6: Solve for x
After substituting and simplifying, you will find an expression for \( x \) in terms of \( t \). The resulting equation will allow you to express \( x \) as:
\[ x = 2 + \frac{2t^2}{1 + t^2} \]

**Hint:** Look for patterns in the resulting equation to isolate \( x \).

### Step 7: Find the Locus
As \( t \) varies, we need to eliminate \( t \) from the equations. Notice that:
\[ \frac{2t^2}{1 + t^2} = 2 - \frac{2}{1 + t^2} \]
This indicates that the locus of the intersection points is a vertical line at \( x = 2 \).

**Hint:** Use the identity \( 1 + t^2 \) to express \( t \) in terms of \( x \) and eliminate it.

### Conclusion
The locus of the point of intersection of any tangent to the parabola \( y^2 = 4a(x - 2) \) with a line perpendicular to it and passing through the focus is:
\[ \boxed{x = 2} \]