Find the locus of the midpoints of the portion of the normal to the parabola `y^2=4a x` intercepted between the curve and the axis.

Explanation

To find the locus of the midpoints of the portion of the normal to the parabola \( y^2 = 4ax \) intercepted between the curve and the axis, we will follow these steps:

### Step 1: Identify the point on the parabola
Let the point \( P \) on the parabola be represented as \( P(at^2, 2at) \), where \( t \) is a parameter.

**Hint:** The coordinates of any point on the parabola \( y^2 = 4ax \) can be expressed in terms of the parameter \( t \).

### Step 2: Write the equation of the normal at point \( P \)
The equation of the normal to the parabola at point \( P(at^2, 2at) \) is given by:
\[ y = -tx + 2at + at^3 \]

**Hint:** The slope of the normal is the negative reciprocal of the slope of the tangent at point \( P \).

### Step 3: Find the x-intercept of the normal
To find the x-intercept \( Q \) of the normal, set \( y = 0 \):
\[ 0 = -tx + 2at + at^3 \]
Rearranging gives:
\[ tx = 2at + at^3 \]
Thus,
\[ x = \frac{2a + at^2}{t} \]
So the coordinates of point \( Q \) are:
\[ Q\left(\frac{2a + at^2}{t}, 0\right) \]

**Hint:** The x-coordinate of the intercept can be found by substituting \( y = 0 \) into the normal equation.

### Step 4: Find the midpoint \( M \) of segment \( PQ \)
The midpoint \( M \) of segment \( PQ \) is given by:
\[ M\left(\frac{x_P + x_Q}{2}, \frac{y_P + y_Q}{2}\right) \]
Substituting the coordinates of \( P \) and \( Q \):
\[ M\left(\frac{at^2 + \frac{2a + at^2}{t}}{2}, \frac{2at + 0}{2}\right) \]
Calculating the x-coordinate:
\[ x_M = \frac{at^2 + \frac{2a + at^2}{t}}{2} = \frac{at^2 + \frac{2a}{t} + at}{2} \]
This simplifies to:
\[ x_M = \frac{a(t^2 + 2/t + t)}{2} \]
Calculating the y-coordinate:
\[ y_M = at \]

**Hint:** The midpoint formula averages the coordinates of the endpoints.

### Step 5: Express \( t \) in terms of \( y_M \)
From \( y_M = at \), we can express \( t \) as:
\[ t = \frac{y_M}{a} \]

**Hint:** Isolate \( t \) to express it in terms of \( y \).

### Step 6: Substitute \( t \) back into the expression for \( x_M \)
Substituting \( t = \frac{y}{a} \) into the expression for \( x_M \):
\[ x_M = \frac{a\left(\left(\frac{y}{a}\right)^2 + 2\left(\frac{y}{a}\right) + \frac{y}{a}\right)}{2} \]
This simplifies to:
\[ x_M = \frac{a\left(\frac{y^2}{a^2} + \frac{2y}{a} + \frac{y}{a}\right)}{2} = \frac{y^2 + 3ay}{2a} \]

**Hint:** Substitute \( t \) into the expression to eliminate the parameter.

### Step 7: Rearranging to find the locus
Rearranging gives:
\[ 2ax_M = y^2 + 3ay \]
Thus, the equation of the locus is:
\[ y^2 + 3ay - 2ax = 0 \]

**Final Answer:** The locus of the midpoints of the portion of the normal to the parabola \( y^2 = 4ax \) intercepted between the curve and the axis is:
\[ y^2 + 3ay - 2ax = 0 \]