If the normals any point to the parabola `x^(2)=4y` cuts the line y = 2 in points whose abscissar are in A.P., them the slopes of the tangents at the 3 conormal points are in

Options

A. AP
B. GP
C. HP
D. None of these

Answer

To solve the problem, we need to find the slopes of the tangents at three co-normal points on the parabola \( x^2 = 4y \) given that the normals at these points intersect the line \( y = 2 \) at points whose abscissas are in arithmetic progression (AP).

### Step 1: Identify the parabola and the normal equation
The given parabola is \( x^2 = 4y \). The parametric equations for points on the parabola can be expressed as:
- \( x = 2t \)
- \( y = t^2 \)

The slope of the tangent at any point \( (2t, t^2) \) on the parabola is given by:
\[ \text{slope} = \frac{dy}{dx} = \frac{2t}{2} = t \]

The equation of the normal at this point is:
\[ y - t^2 = -\frac{1}{t}(x - 2t) \]
Rearranging gives:
\[ y = -\frac{1}{t}x + 2 + t^2 \]

### Step 2: Find where the normal intersects the line \( y = 2 \)
Setting \( y = 2 \) in the normal equation:
\[ 2 = -\frac{1}{t}x + 2 + t^2 \]
This simplifies to:
\[ -\frac{1}{t}x + t^2 = 0 \implies x = t^3 \]

### Step 3: Determine the abscissas of the intersection points
Let the three points where the normals intersect \( y = 2 \) correspond to parameters \( t_1, t_2, t_3 \):
- The abscissas are \( t_1^3, t_2^3, t_3^3 \).

According to the problem, these abscissas are in AP:
\[ t_1^3, t_2^3, t_3^3 \text{ are in AP} \]

### Step 4: Use the property of AP
For three numbers \( a, b, c \) to be in AP, the condition is:
\[ 2b = a + c \]
Applying this to our case:
\[ 2t_2^3 = t_1^3 + t_3^3 \]

### Step 5: Relate the parameters using the property of cubes
Using the identity for cubes:
\[ a^3 + c^3 = (a + c)(a^2 - ac + c^2) \]
We can express \( t_1^3 + t_3^3 \) in terms of \( t_2 \):
\[ t_1^3 + t_3^3 = (t_1 + t_3)((t_1^2 + t_3^2) - t_1t_3) \]
Since \( t_1 + t_3 = 2t_2 \) (from the AP condition), we can substitute and simplify.

### Step 6: Find the slopes of the tangents at the co-normal points
The slopes of the tangents at points \( t_1, t_2, t_3 \) are:
- \( m_1 = t_1 \)
- \( m_2 = t_2 \)
- \( m_3 = t_3 \)

### Step 7: Determine the relationship of the slopes
Since \( t_1, t_2, t_3 \) are in AP, their slopes will also have a specific relationship. The slopes \( m_1, m_2, m_3 \) will be in harmonic progression (HP) because the points are co-normal.

### Conclusion
Thus, the slopes of the tangents at the three co-normal points are in harmonic progression (HP).

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