If a tangent to the parabola `y^2 = 4ax` meets the axis of the parabola in `T` and the tangent at the vertex `A` in `Y,` and the rectangle `TAYG` is completed, show that the locus of `G` is `y^2 + ax = 0.`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Parabola and Tangent
The equation of the parabola is given as \( y^2 = 4ax \). The vertex of the parabola is at the origin \( A(0, 0) \).

### Step 2: Equation of the Tangent
The equation of the tangent to the parabola at any point \( (at^2, 2at) \) is given by:
\[ ty = x + at^2 \]
This is the general equation of the tangent line.

### Step 3: Find Point T
Point \( T \) is where the tangent meets the x-axis. Since the x-axis has \( y = 0 \), we substitute \( y = 0 \) into the tangent equation:
\[ t(0) = x + at^2 \implies x = -at^2 \]
Thus, the coordinates of point \( T \) are \( (-at^2, 0) \).

### Step 4: Find Point Y
Point \( Y \) is where the tangent at the vertex \( A \) meets the tangent line. The tangent at the vertex \( A \) is the y-axis, which has the equation \( x = 0 \). We substitute \( x = 0 \) into the tangent equation:
\[ ty = 0 + at^2 \implies y = \frac{at^2}{t} = at \]
Thus, the coordinates of point \( Y \) are \( (0, at) \).

### Step 5: Coordinates of Point G
Let point \( G \) have coordinates \( (h, k) \). The rectangle \( TAYG \) implies that:
- The x-coordinate of \( G \) is the same as that of \( T \), hence \( h = -at^2 \).
- The y-coordinate of \( G \) is the same as that of \( Y \), hence \( k = at \).

### Step 6: Eliminate t
From the equations \( h = -at^2 \) and \( k = at \), we can express \( t \) in terms of \( k \):
\[ t = \frac{k}{a} \]
Now substituting \( t \) back into the equation for \( h \):
\[ h = -a\left(\frac{k}{a}\right)^2 = -\frac{k^2}{a} \]
Rearranging gives:
\[ ah + k^2 = 0 \]

### Step 7: Locus of Point G
The equation \( ah + k^2 = 0 \) can be rewritten as:
\[ k^2 + ah = 0 \]
Substituting back \( h \) and \( k \) with \( x \) and \( y \) respectively gives us:
\[ y^2 + ax = 0 \]
This is the locus of point \( G \).

### Conclusion
Thus, we have shown that the locus of point \( G \) is:
\[ y^2 + ax = 0 \]