Let tangent PQ and PR are drawn from the point `P(-2, 4)` to the parabola `y^(2)=4x`. If S is the focus of the parabola `y^(2)=4x`, then the value (in units) of `RS+SQ` is equal to

Answer

To solve the problem step-by-step, we will follow the mathematical principles related to parabolas and tangents.

### Step 1: Identify the focus of the parabola
The given parabola is \(y^2 = 4x\). The standard form of a parabola is \(y^2 = 4ax\), where the focus is located at \((a, 0)\). Here, \(4a = 4\), which gives \(a = 1\). Thus, the focus \(S\) of the parabola is at the point:
\[ S(1, 0) \]

**Hint:** Remember that for the parabola \(y^2 = 4ax\), the focus is at \((a, 0)\).

### Step 2: Define the points of tangency
Let the points of tangency from point \(P(-2, 4)\) to the parabola be \(Q(t_1)\) and \(R(t_2)\). The coordinates of these points on the parabola can be expressed as:
\[ Q(t_1) = (t_1^2, 2t_1) \quad \text{and} \quad R(t_2) = (t_2^2, 2t_2) \]

**Hint:** The points on the parabola can be parameterized using \(t\) values.

### Step 3: Set up the equations based on the point \(P\)
Since \(P(-2, 4)\) is an external point from which tangents are drawn, we can use the properties of tangents to the parabola. The conditions for the tangents from point \(P\) give us:
1. \(t_1 + t_2 = 4\) (sum of the roots)
2. \(t_1 t_2 = -2\) (product of the roots)

**Hint:** Use the relationships between the coordinates of point \(P\) and the parameters \(t_1\) and \(t_2\).

### Step 4: Calculate the distances \(RS\) and \(SQ\)
Now, we need to find the distances \(RS\) and \(SQ\).

1. **Distance \(RS\)**:
\[ RS = \sqrt{(t_2^2 - 1)^2 + (2t_2 - 0)^2} = \sqrt{(t_2^2 - 1)^2 + (2t_2)^2} \]
Simplifying this:
\[ RS = \sqrt{(t_2^2 - 1)^2 + 4t_2^2} = \sqrt{t_2^4 - 2t_2^2 + 1 + 4t_2^2} = \sqrt{t_2^4 + 2t_2^2 + 1} = \sqrt{(t_2^2 + 1)^2} = t_2^2 + 1 \]

2. **Distance \(SQ\)**:
\[ SQ = \sqrt{(t_1^2 - 1)^2 + (2t_1 - 0)^2} = \sqrt{(t_1^2 - 1)^2 + (2t_1)^2} \]
Similarly, simplifying gives:
\[ SQ = \sqrt{(t_1^2 - 1)^2 + 4t_1^2} = t_1^2 + 1 \]

### Step 5: Sum the distances \(RS + SQ\)
Now, we can find \(RS + SQ\):
\[ RS + SQ = (t_2^2 + 1) + (t_1^2 + 1) = t_1^2 + t_2^2 + 2 \]

Using the identity \(t_1^2 + t_2^2 = (t_1 + t_2)^2 - 2t_1t_2\):
\[ t_1^2 + t_2^2 = 4^2 - 2(-2) = 16 + 4 = 20 \]
Thus,
\[ RS + SQ = 20 + 2 = 22 \]

### Final Answer
The value of \(RS + SQ\) is:
\[ \boxed{22} \]