To solve the problem, we need to find the relationship between the lengths \( SP \), \( SP' \), and \( SL \) for the focal chord \( PSP' \) of the parabola \( y^2 = 4ax \) and the semi-latus rectum \( SL \).
### Step 1: Identify the points on the parabola
The parabola is given by the equation \( y^2 = 4ax \). The focus \( S \) of the parabola is located at the point \( (a, 0) \).
Let \( P \) and \( P' \) be points on the focal chord. The coordinates of point \( P \) can be expressed as:
\[
P(t) = (at^2, 2at)
\]
And the coordinates of point \( P' \) can be expressed as:
\[
P'(-t) = \left( a\frac{1}{t^2}, -2a\frac{1}{t} \right)
\]
### Step 2: Calculate the lengths \( SP \) and \( SP' \)
To find the lengths \( SP \) and \( SP' \), we will use the distance formula.
**Length \( SP \):**
\[
SP = \sqrt{(a - at^2)^2 + (0 - 2at)^2}
\]
Calculating this gives:
\[
= \sqrt{(a(1 - t^2))^2 + (2at)^2}
\]
\[
= \sqrt{a^2(1 - 2t^2 + t^4) + 4a^2t^2}
\]
\[
= \sqrt{a^2(1 + t^4 + 2t^2)} = a\sqrt{(1 + t^2)^2} = a(1 + t^2)
\]
**Length \( SP' \):**
\[
SP' = \sqrt{(a - \frac{a}{t^2})^2 + (0 + 2a\frac{1}{t})^2}
\]
Calculating this gives:
\[
= \sqrt{(a(1 - \frac{1}{t^2}))^2 + (2a\frac{1}{t})^2}
\]
\[
= \sqrt{a^2(1 - \frac{2}{t^2} + \frac{1}{t^4}) + \frac{4a^2}{t^2}}
\]
\[
= \sqrt{a^2(1 + \frac{1}{t^4} + \frac{2}{t^2})} = a(1 + \frac{1}{t^2})
\]
### Step 3: Calculate the length of the semi-latus rectum \( SL \)
The semi-latus rectum \( SL \) is given by:
\[
SL = \frac{4a}{2} = 2a
\]
### Step 4: Check the relationship between \( SP \), \( SP' \), and \( SL \)
Now we have:
- \( SP = a(1 + t^2) \)
- \( SP' = a(1 + \frac{1}{t^2}) \)
- \( SL = 2a \)
To check if \( SP \), \( SP' \), and \( SL \) are in Harmonic Progression (HP), we need to verify:
\[
\frac{1}{SP} + \frac{1}{SP'} = \frac{2}{SL}
\]
Calculating the left-hand side:
\[
\frac{1}{SP} = \frac{1}{a(1 + t^2)}, \quad \frac{1}{SP'} = \frac{1}{a(1 + \frac{1}{t^2})}
\]
\[
\Rightarrow \frac{1}{SP} + \frac{1}{SP'} = \frac{1}{a(1 + t^2)} + \frac{1}{a(1 + \frac{1}{t^2})}
\]
Finding a common denominator:
\[
= \frac{(1 + \frac{1}{t^2}) + (1 + t^2)}{a(1 + t^2)(1 + \frac{1}{t^2})}
\]
\[
= \frac{2 + t^2 + \frac{1}{t^2}}{a(1 + t^2)(1 + \frac{1}{t^2})}
\]
Now calculating the right-hand side:
\[
\frac{2}{SL} = \frac{2}{2a} = \frac{1}{a}
\]
### Step 5: Equate both sides
To show they are equal:
\[
\frac{2 + t^2 + \frac{1}{t^2}}{(1 + t^2)(1 + \frac{1}{t^2})} = 1
\]
This simplifies to:
\[
2 + t^2 + \frac{1}{t^2} = (1 + t^2)(1 + \frac{1}{t^2}) = 1 + t^2 + \frac{1}{t^2} + 1
\]
Thus, we confirm that:
\[
SP, SP', SL \text{ are in Harmonic Progression (HP)}.
\]
### Final Answer
Therefore, \( SP \), \( SP' \), and \( SL \) are in Harmonic Progression (HP).
