The set of points on the axis of the parabola `(x-1)^(2)=8(y+2)` from where three distinct normals can be drawn to the parabola is the set (h,k) of points satisfying

Options

A. `hgt2`
B. `hgt1`
C. `kgt2`
D. none of these

Explanation

To solve the problem of finding the set of points on the axis of the parabola \((x-1)^2 = 8(y+2)\) from which three distinct normals can be drawn, we can follow these steps:

### Step 1: Identify the Parabola's Vertex and Parameters
The given equation of the parabola is \((x-1)^2 = 8(y+2)\). This is in the standard form \((x-h)^2 = 4p(y-k)\), where:
- The vertex \((h, k)\) is \((1, -2)\).
- Here, \(4p = 8\), so \(p = 2\).

### Step 2: Determine the Focus and Directrix
For the parabola, the focus is located at \((h, k + p) = (1, -2 + 2) = (1, 0)\), and the directrix is the line \(y = k - p = -2 - 2 = -4\).

### Step 3: Find the Condition for Three Normals
For any parabola, the condition to draw three distinct normals from a point on the axis is that the distance from the vertex must be greater than \(2p\). In this case, since \(p = 2\), we have:
\[ 2p = 2 \times 2 = 4. \]
Thus, the distance from the vertex (which is at \(y = -2\)) to the point \((h, k)\) must satisfy:
\[ k > -2 + 4 \quad \Rightarrow \quad k > 2. \]

### Step 4: Conclusion
The set of points \((h, k)\) on the axis of the parabola from which three distinct normals can be drawn is given by:
\[ k > 2. \]
Thus, the answer is the set of points \((h, k)\) such that \(k > 2\).

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