If the tangents at the points `Pa n dQ` on the parabola `y^2=4a x` meet at `T ,a n dS` is its focus, the prove that `S P ,S T ,a n dS Q` are in GP.

Explanation

To prove that the distances \( SP, ST, \) and \( SQ \) are in geometric progression (GP), we will follow these steps:

### Step 1: Identify the focus and points on the parabola
The equation of the parabola is given by \( y^2 = 4ax \). The focus \( S \) of this parabola is at the point \( (a, 0) \). Let points \( P \) and \( Q \) on the parabola have parameters \( t_1 \) and \( t_2 \) respectively. The coordinates of these points are:
- \( P(a t_1^2, 2a t_1) \)
- \( Q(a t_2^2, 2a t_2) \)

### Step 2: Find the equations of the tangents at points \( P \) and \( Q \)
The equation of the tangent to the parabola \( y^2 = 4ax \) at point \( (at^2, 2at) \) is given by:
\[ yy_1 = 2a(x + x_1) \]
For point \( P \) (with parameter \( t_1 \)):
\[ y(2a t_1) = 2a(x + a t_1^2) \]
This simplifies to:
\[ y = \frac{2a}{2a t_1}(x + a t_1^2) = \frac{1}{t_1}(x + a t_1^2) \]

For point \( Q \) (with parameter \( t_2 \)):
\[ y(2a t_2) = 2a(x + a t_2^2) \]
This simplifies to:
\[ y = \frac{2a}{2a t_2}(x + a t_2^2) = \frac{1}{t_2}(x + a t_2^2) \]

### Step 3: Find the point of intersection \( T \) of the tangents
To find the coordinates of point \( T \), we need to solve the equations of the tangents simultaneously. Setting the two equations equal gives:
\[ \frac{1}{t_1}(x + a t_1^2) = \frac{1}{t_2}(x + a t_2^2) \]
Cross-multiplying and simplifying leads to:
\[ t_2(x + a t_1^2) = t_1(x + a t_2^2) \]
Rearranging gives:
\[ (t_2 - t_1)x = a(t_1 t_2^2 - t_2 t_1^2) \]
Thus, the x-coordinate of \( T \) is:
\[ x_T = \frac{a(t_1 t_2^2 - t_2 t_1^2)}{t_2 - t_1} \]
The y-coordinate can be found by substituting \( x_T \) back into either tangent equation.

### Step 4: Calculate distances \( SP, ST, \) and \( SQ \)
1. **Distance \( SP \)**:
\[ SP = \sqrt{(x_P - a)^2 + (y_P - 0)^2} = \sqrt{(at_1^2 - a)^2 + (2at_1)^2} \]
Simplifying gives:
\[ SP = a\sqrt{(t_1^2 - 1)^2 + 4t_1^2} = a\sqrt{t_1^4 + 2t_1^2 + 1} = a\sqrt{(t_1^2 + 1)^2} = a(t_1^2 + 1) \]

2. **Distance \( SQ \)**:
Similarly, we find:
\[ SQ = a(t_2^2 + 1) \]

3. **Distance \( ST \)**:
Using the coordinates of \( T \) and \( S \):
\[ ST = \sqrt{(x_T - a)^2 + (y_T - 0)^2} \]
After simplification, we find:
\[ ST = a\sqrt{(1 + t_1^2)(1 + t_2^2)} \]

### Step 5: Show that \( SP, ST, SQ \) are in GP
To show that \( SP, ST, SQ \) are in GP, we need to prove:
\[ ST^2 = SP \cdot SQ \]
Calculating:
\[ ST^2 = a^2(1 + t_1^2)(1 + t_2^2) \]
And:
\[ SP \cdot SQ = a(t_1^2 + 1) \cdot a(t_2^2 + 1) = a^2(1 + t_1^2)(1 + t_2^2) \]
Since both expressions are equal, we conclude that \( SP, ST, SQ \) are in GP.