Consider the parabola `x^(2) +4y = 0`. Let `P(a,b)` be any fixed point inside the parabola and let S be the focus of the parabola. Then the minimum value at `SQ +PQ` as point Q moves on the parabola is (a) `|1 -a|` (b) `|ab|+1` (c) `sqrt(a^(2)+b^(2))` (d) `1-b`

Options

A. `|1 -a|`
B. `|ab|+1`
C. `sqrt(a^(2)+b^(2))`
D. `1-b`

Explanation

To solve the problem, we need to analyze the given parabola and the points involved. Let's go through the solution step by step.

### Step 1: Identify the Parabola and its Focus
The equation of the parabola is given as:
\[ x^2 + 4y = 0 \]
This can be rewritten as:
\[ y = -\frac{1}{4}x^2 \]
This is a downward-opening parabola. The standard form of a parabola that opens downwards is:
\[ y = -\frac{1}{4p}x^2 \]
From this, we can identify that \( p = 1 \). The focus of the parabola is located at:
\[ (0, p) = (0, 1) \]

### Step 2: Determine the Directrix
The directrix of the parabola is given by the equation:
\[ y = -p = -1 \]

### Step 3: Define Points
Let \( P(a, b) \) be a fixed point inside the parabola. Since \( P \) is inside the parabola, it satisfies:
\[ b > -\frac{1}{4}a^2 \]

Let \( S(0, 1) \) be the focus of the parabola, and let \( Q(x, y) \) be any point on the parabola. Since \( Q \) lies on the parabola, it also satisfies:
\[ y = -\frac{1}{4}x^2 \]

### Step 4: Express the Distances
We need to find the minimum value of \( SQ + PQ \):
- The distance \( SQ \) from the focus \( S(0, 1) \) to point \( Q(x, y) \) is given by:
\[ SQ = \sqrt{(x - 0)^2 + (y - 1)^2} = \sqrt{x^2 + \left(-\frac{1}{4}x^2 - 1\right)^2} \]
- The distance \( PQ \) from point \( P(a, b) \) to point \( Q(x, y) \) is given by:
\[ PQ = \sqrt{(x - a)^2 + \left(-\frac{1}{4}x^2 - b\right)^2} \]

### Step 5: Minimize the Total Distance
To minimize \( SQ + PQ \), we can use the geometric property that the sum of distances from a point to a focus and a point on the directrix is minimized when the point lies on the line segment connecting the focus and the foot of the perpendicular from the point to the directrix.

Let \( N(0, -1) \) be the foot of the perpendicular from \( P \) to the directrix. The minimum value occurs when \( P, N, Q \) are collinear.

### Step 6: Calculate the Minimum Value
The minimum value of \( SQ + PQ \) can be expressed as:
\[ SQ + PQ = PN \]
The distance \( PN \) can be calculated as:
\[ PN = \sqrt{(0 - a)^2 + (-1 - b)^2} = \sqrt{a^2 + (b + 1)^2} \]

### Step 7: Analyze the Result
Since we need to find the minimum value of \( SQ + PQ \), we can observe that:
- The minimum occurs when \( Q \) is directly below \( P \) at the point where the line connecting \( S \) and \( N \) intersects the parabola.

Thus, the minimum value of \( SQ + PQ \) is:
\[ 1 - b \]

### Final Answer
The minimum value of \( SQ + PQ \) as point \( Q \) moves on the parabola is:
\[ \boxed{1 - b} \]