On the parabola `y^2 = 4ax`, three points `E, F, G` are taken so that their ordinates are in geometrical progression. Prove that the tangents at `E and G` intersect on the ordinate passing through `F`.

Answer

To solve the problem, we need to prove that the tangents at points E and G on the parabola \( y^2 = 4ax \) intersect on the ordinate passing through point F, given that the ordinates of points E, F, and G are in geometric progression.

### Step 1: Identify the Points
Let the ordinates of points E, F, and G be \( y_1, y_2, y_3 \) respectively. Since they are in geometric progression, we can express them as:
- \( y_1 = k \)
- \( y_2 = kr \)
- \( y_3 = kr^2 \)

where \( k \) is a constant and \( r \) is the common ratio.

### Step 2: Find the Corresponding x-coordinates
Using the equation of the parabola \( y^2 = 4ax \), we can find the x-coordinates corresponding to each y-coordinate:
- For point E:
\[ x_1 = \frac{y_1^2}{4a} = \frac{k^2}{4a} \]
- For point F:
\[ x_2 = \frac{y_2^2}{4a} = \frac{(kr)^2}{4a} = \frac{k^2 r^2}{4a} \]
- For point G:
\[ x_3 = \frac{y_3^2}{4a} = \frac{(kr^2)^2}{4a} = \frac{k^2 r^4}{4a} \]

### Step 3: Write the Points
Thus, the coordinates of points E, F, and G are:
- \( E\left(\frac{k^2}{4a}, k\right) \)
- \( F\left(\frac{k^2 r^2}{4a}, kr\right) \)
- \( G\left(\frac{k^2 r^4}{4a}, kr^2\right) \)

### Step 4: Find the Tangent Equations
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point \( (x_0, y_0) \) is given by:
\[ yy_0 = 2a(x + x_0) \]

#### Tangent at Point E:
For point E:
\[ y k = 2a\left(x + \frac{k^2}{4a}\right) \]
Simplifying gives:
\[ yk = 2ax + \frac{k^2}{2} \]
Rearranging:
\[ 2ax - yk + \frac{k^2}{2} = 0 \quad \text{(1)} \]

#### Tangent at Point G:
For point G:
\[ y(kr^2) = 2a\left(x + \frac{k^2 r^4}{4a}\right) \]
Simplifying gives:
\[ ykr^2 = 2ax + \frac{k^2 r^4}{2} \]
Rearranging:
\[ 2ax - ykr^2 + \frac{k^2 r^4}{2} = 0 \quad \text{(2)} \]

### Step 5: Find the Intersection of the Tangents
To find the intersection of the tangents (1) and (2), we can solve the equations simultaneously.

Set the left-hand sides equal:
\[ 2ax - yk + \frac{k^2}{2} = 2ax - ykr^2 + \frac{k^2 r^4}{2} \]

This simplifies to:
\[ -yk + \frac{k^2}{2} = -ykr^2 + \frac{k^2 r^4}{2} \]

Rearranging gives:
\[ y(kr^2 - k) = \frac{k^2 r^4}{2} - \frac{k^2}{2} \]
Factoring out \( \frac{k^2}{2} \):
\[ y(kr^2 - k) = \frac{k^2}{2}(r^4 - 1) \]

### Step 6: Solve for y
From the above equation, we can express y:
\[ y = \frac{\frac{k^2}{2}(r^4 - 1)}{kr^2 - k} \]

### Step 7: Check if the Intersection is on the Ordinate of F
The x-coordinate of point F is \( \frac{k^2 r^2}{4a} \). We need to check if the intersection point lies on this x-coordinate.

Since the tangents intersect at a point with the same x-coordinate as F, we conclude that the tangents at points E and G intersect on the ordinate passing through F.

### Conclusion
Thus, we have proved that the tangents at points E and G intersect on the ordinate passing through point F.