the distance between point P which is on parabola y^2=8x and the focus F is 5, then the x-coordinate of point P is

Given the problem, we have a point \( P(x, y) \) on the parabola \( y^2 = 8x \), and the distance between this point \( P \) and the focus \( F \) is 5.

### Step 1: Identify the focus of the parabola
The equation of the parabola is \( y^2 = 8x \). The general form of a parabola that opens to the right is \( y^2 = 4ax \). By comparing \( y^2 = 4ax \) with \( y^2 = 8x \), we find:
\[
4a = 8 \implies a = 2
\]
The focus \( F \) of the parabola is located at \( (a, 0) = (2, 0) \).

### Step 2: Use the distance formula
The distance between the point \( P(x, y) \) and the focus \( F(2, 0) \) is given as 5. Using the distance formula:
\[
\sqrt{(x - 2)^2 + (y - 0)^2} = 5
\]
Square both sides to eliminate the square root:
\[
(x - 2)^2 + y^2 = 25
\]

### Step 3: Substitute the parabola's equation
Since \( y^2 = 8x \), we can substitute \( y^2 \) in the equation:
\[
(x - 2)^2 + 8x = 25
\]
Expand and simplify:
\[
x^2 - 4x + 4 + 8x = 25
\]
\[
x^2 + 4x + 4 = 25
\]
\[
x^2 + 4x - 21 = 0
\]

### Step 4: Solve the quadratic equation
Solve the quadratic equation \( x^2 + 4x - 21 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = 4 \), and \( c = -21 \). Substitute these values:
\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-21)}}{2(1)}
\]
\[
x = \frac{-4 \pm \sqrt{16 + 84}}{2}
\]
\[
x = \frac{-4 \pm \sqrt{100}}{2}
\]
\[
x = \frac{-4 \pm 10}{2}
\]
\[
x = 3 \quad \text{or} \quad x = -7
\]

### Conclusion
The possible x-coordinates of point \( P \) are \( \boxed{3} \) and \( \boxed{-7} \).

Would you like more details or have any other questions?

### Follow-up Questions:
1. What is the significance of the focus in a parabola?
2. How do you find the focus of a parabola given its equation?
3. Can a point on a parabola be closer to the vertex than the focus?
4. How would the calculation change if the parabola opened upwards instead of to the right?
5. What is the geometric meaning of the directrix in relation to the focus?

### Tip:
Always verify your quadratic solutions by substituting back into the original equation to ensure correctness.

Learn how to find the x-coordinate of point P on the parabola y^2 = 8x given that its distance to the focus F is 5. This problem involves using the distance formula and solving a quadratic equation derived from the parabola's equation. Suitable for high school students (Grades 10-12).

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