Tangent to the parabola `y^2 = 4ax` is of the form `y = mx + a/m` where m is the slope of the tangent.

Answer

To show that the tangent to the parabola \( y^2 = 4ax \) is of the form \( y = mx + \frac{a}{m} \), we will follow these steps:

### Step 1: Find the derivative of the parabola
The equation of the parabola is given by:
\[ y^2 = 4ax \]
To find the slope of the tangent line, we differentiate both sides with respect to \( x \).
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \]
Using the chain rule on the left side, we have:
\[ 2y \frac{dy}{dx} = 4a \]
Now, solving for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

**Hint:** Remember that the derivative gives the slope of the tangent line at any point on the curve.

### Step 2: Identify the point of tangency
Let the point of tangency on the parabola be \( (at^2, 2at) \). Here, \( t \) is a parameter that represents the point on the parabola.

**Hint:** The coordinates \( (at^2, 2at) \) are derived from the parametric equations of the parabola.

### Step 3: Substitute the point into the slope formula
Using the point \( (at^2, 2at) \), we can find the slope \( m \) of the tangent line at this point:
\[ m = \frac{2a}{2at} = \frac{1}{t} \]

**Hint:** The slope \( m \) is derived from substituting the coordinates of the point of tangency into the derivative.

### Step 4: Write the equation of the tangent line
Using the point-slope form of the equation of a line, we have:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (at^2, 2at) \) and \( m = \frac{1}{t} \):
\[ y - 2at = \frac{1}{t}(x - at^2) \]

**Hint:** The point-slope form is a useful way to express the equation of a line given a point and a slope.

### Step 5: Rearranging the equation
Rearranging the equation gives:
\[ y - 2at = \frac{x}{t} - a t \]
\[ y = \frac{x}{t} + 2at - at \]
\[ y = \frac{x}{t} + a\left(\frac{1}{t}\right) \]

**Hint:** Pay attention to how terms are combined and rearranged to isolate \( y \).

### Step 6: Final form of the tangent equation
Now, we can rewrite the equation as:
\[ y = mx + \frac{a}{m} \]
where \( m = \frac{1}{t} \).

**Hint:** This final form matches the required format, confirming that the tangent line to the parabola is indeed of the specified form.

### Conclusion
Thus, we have shown that the tangent to the parabola \( y^2 = 4ax \) is of the form \( y = mx + \frac{a}{m} \), where \( m \) is the slope of the tangent.