For what values of 'a' will the tangents drawn to the parabola `y^2 = 4ax` from a point, not on the y-axis, will be normal to the parabola `x^2 = 4y.`

Answer

To find the values of 'a' for which the tangents drawn to the parabola \( y^2 = 4ax \) from a point not on the y-axis are normal to the parabola \( x^2 = 4y \), we can follow these steps:

### Step 1: Identify Points on the Parabola \( x^2 = 4y \)
A point on the parabola \( x^2 = 4y \) can be represented as \( (2t, t^2) \).

### Step 2: Find the Slope of the Tangent to the Parabola \( y^2 = 4ax \)
The slope of the tangent to the parabola \( y^2 = 4ax \) at a point \( (x_1, y_1) \) can be derived from the equation:
\[ y = mx + \frac{a}{m} \]
For the point \( (2t, t^2) \), we can find the slope by differentiating \( y^2 = 4ax \):
\[ 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \]
Substituting \( y = t^2 \):
\[ \frac{dy}{dx} = \frac{2a}{t^2} \]

### Step 3: Equation of the Tangent Line
Using the point-slope form of the equation of a line, the equation of the tangent line at the point \( (2t, t^2) \) is:
\[ y - t^2 = \frac{2a}{t^2}(x - 2t) \]

### Step 4: Find the Slope of the Normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ m_{normal} = -\frac{t^2}{2a} \]

### Step 5: Equation of the Normal Line
The equation of the normal line at the point \( (2t, t^2) \) is:
\[ y - t^2 = -\frac{t^2}{2a}(x - 2t) \]

### Step 6: Find the Condition for Normality
For the normal line to be normal to the parabola \( x^2 = 4y \), we need to find the condition that relates the slopes. The slope of the tangent to the parabola \( x^2 = 4y \) at any point can be given as:
\[ m_{tangent} = \frac{x}{2} \]
Substituting \( x = 2t \):
\[ m_{tangent} = t \]

### Step 7: Set the Condition for Normality
For the normal to the first parabola to be tangent to the second, we set:
\[ -\frac{t^2}{2a} \cdot t = -1 \]
This simplifies to:
\[ \frac{t^3}{2a} = 1 \implies t^3 = 2a \]

### Step 8: Substitute Back to Find 'a'
Substituting \( t^3 = 2a \) back into the equation for the normal gives us a quadratic in \( t \):
\[ t^2 + \frac{2a}{t} + 2 = 0 \]
The discriminant of this quadratic must be greater than or equal to zero for real roots:
\[ \left(\frac{2a}{t}\right)^2 - 4 \cdot 1 \cdot 2 \geq 0 \]
This leads to:
\[ 4a^2 - 8t \geq 0 \implies a^2 \geq 2t \]

### Step 9: Solve for 'a'
From the previous steps, we find that:
\[ a^2 > 8 \implies |a| > 2\sqrt{2} \]
Thus, the values of \( a \) are:
\[ a < -2\sqrt{2} \quad \text{or} \quad a > 2\sqrt{2} \]

### Final Answer
The values of \( a \) for which the tangents drawn to the parabola \( y^2 = 4ax \) from a point not on the y-axis are normal to the parabola \( x^2 = 4y \) are:
\[ a < -2\sqrt{2} \quad \text{or} \quad a > 2\sqrt{2} \]