The locus of the mid-points of the portion of the normal to the parabola `y^2 = 4ax` intercepted between the curve and the axis is another parabola

Options

A. `y^2 = 2a (x-a)`
B. `y^2 = a (x+a)`
C. `y^2 = a(x-a)`
D. none

Answer

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 x-axis, we can follow these steps:

### Step 1: Understand the Parabola
The given parabola is \( y^2 = 4ax \). The vertex of this parabola is at the origin (0,0), and it opens to the right.

**Hint:** Remember that the standard form of a parabola \( y^2 = 4ax \) indicates its orientation and vertex.

### Step 2: Parametric Coordinates
Let the point \( P \) on the parabola be given by the parametric coordinates:
\[ P(t) = (at^2, 2at) \]
where \( t \) is a parameter.

**Hint:** Use the parametric equations of the parabola to express points on it.

### Step 3: Equation of the Normal
The equation of the normal to the parabola at point \( P(t) \) is given by:
\[ y = -tx + 2a + 2at \]
This normal line will intersect the x-axis where \( y = 0 \).

**Hint:** The slope of the normal is the negative reciprocal of the slope of the tangent at that point.

### Step 4: Find the Intersection with the X-axis
Set \( y = 0 \) in the normal equation:
\[ 0 = -tx + 2a + 2at \]
Solving for \( x \):
\[ tx = 2a + 2at \implies x = \frac{2a(1 + t)}{t} \]
Thus, the coordinates of point \( Q \) where the normal intersects the x-axis are:
\[ Q\left(\frac{2a(1 + t)}{t}, 0\right) \]

**Hint:** Substitute \( y = 0 \) to find where the normal intersects the x-axis.

### Step 5: Midpoint Calculation
The midpoint \( M \) of segment \( PQ \) is given by:
\[ M = \left( \frac{at^2 + \frac{2a(1 + t)}{t}}{2}, \frac{2at + 0}{2} \right) \]
This simplifies to:
\[ M = \left( \frac{at^2 + \frac{2a(1 + t)}{t}}{2}, at \right) \]

**Hint:** Use the midpoint formula to find the coordinates of \( M \).

### Step 6: Simplifying the x-coordinate of M
To simplify the x-coordinate:
\[ x = \frac{at^2 + \frac{2a(1 + t)}{t}}{2} = \frac{at^2 + 2a + 2at}{2t} = \frac{a(t^2 + 2t + 2)}{2t} \]
This can be rewritten as:
\[ x = \frac{a(t + 1)^2}{2t} + a \]

**Hint:** Factor and simplify expressions to find relationships between \( x \) and \( t \).

### Step 7: Expressing in terms of \( y \)
From the y-coordinate:
\[ y = at \]
Thus, \( t = \frac{y}{a} \). Substitute \( t \) back into the expression for \( x \):
\[ x = \frac{a\left(\frac{y}{a} + 1\right)^2}{2\left(\frac{y}{a}\right)} + a \]
This simplifies to:
\[ x = \frac{(y + a)^2}{2y} + a \]

**Hint:** Substitute \( t \) in terms of \( y \) to find a relationship between \( x \) and \( y \).

### Step 8: Rearranging to Find the Locus
After simplification, we find that the locus of the midpoints \( M \) is given by:
\[ y^2 = a(x - a) \]
This is the equation of a parabola.

**Hint:** Rearranging the equation will help you identify the standard form of a parabola.

### Conclusion
Thus, the locus of the midpoints of the portion of the normal to the parabola \( y^2 = 4ax \) intercepted between the curve and the x-axis is another parabola given by:
\[ y^2 = a(x - a) \]