AB is a chord of the parabola `y^2 = 4ax` with its vertex at A. BC is drawn perpendicular to AB meeting the axis at C.The projecton of BC on the axis of the parabola is

Options

A. a
B. 2a
C. 4a
D. 8a

Explanation

To solve the problem step by step, we will analyze the given information and apply the relevant concepts of the parabola and geometry.

### Step 1: Understand the Parabola
The equation of the parabola is given as \( y^2 = 4ax \). This parabola opens to the right with its vertex at the origin (0, 0).

**Hint:** Remember that the vertex form of a parabola can help you identify its vertex and orientation.

### Step 2: Identify Points A and B
Since point A is at the vertex, we have:
- \( A(0, 0) \)

Let point B be a point on the parabola. We can express the coordinates of B in terms of a parameter \( t \):
- \( B(at^2, 2at) \)

**Hint:** The coordinates of points on the parabola can be expressed using the parameterization \( (at^2, 2at) \).

### Step 3: Determine the Slope of Line AB
To find the slope of line AB:
- The coordinates of A are \( (0, 0) \) and those of B are \( (at^2, 2at) \).
- The slope of line AB is given by:
\[ \text{slope of AB} = \frac{2at - 0}{at^2 - 0} = \frac{2a t}{at^2} = \frac{2}{t} \]

**Hint:** The slope formula is \( \frac{y_2 - y_1}{x_2 - x_1} \).

### Step 4: Find the Slope of Line BC
Since BC is perpendicular to AB, the slope of BC will be the negative reciprocal of the slope of AB:
\[ \text{slope of BC} = -\frac{t}{2} \]

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

### Step 5: Write the Equation of Line BC
Using point-slope form, the equation of line BC can be written as:
\[ y - 2at = -\frac{t}{2}(x - at^2) \]
Simplifying this, we get:
\[ 2y - 4at = -tx + at^2 \]
Rearranging gives:
\[ tx + 2y = at^2 + 4at \]

**Hint:** The point-slope form is useful for writing the equation of a line when you have a point and a slope.

### Step 6: Find the Intersection Point C
Point C lies on the x-axis, where \( y = 0 \):
\[ t x + 2(0) = at^2 + 4at \implies tx = at^2 + 4at \]
Thus, solving for \( x \):
\[ x = \frac{at^2 + 4at}{t} = at + 4a \]
So, the coordinates of C are:
\[ C(at + 4a, 0) \]

**Hint:** Substitute \( y = 0 \) to find the x-coordinate of point C.

### Step 7: Find the Projection of BC on the x-axis
The projection of BC on the x-axis is the horizontal distance from point M (the foot of the perpendicular from B to the x-axis) to point C. The x-coordinate of M is the same as that of B, which is \( at^2 \).

Thus, the projection length \( MC \) is:
\[ MC = |(at + 4a) - at^2| \]
This simplifies to:
\[ MC = |4a| = 4a \]

**Hint:** The projection length is the absolute difference between the x-coordinates of points C and M.

### Final Answer
The projection of BC on the axis of the parabola is \( 4a \).

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