Find the equations of the tangent and normal to the parabola :
`y^(2)=4ax" at "(at^(2), 2at)`

Answer

To find the equations of the tangent and normal to the parabola \( y^2 = 4ax \) at the point \( (at^2, 2at) \), we can follow these steps:

### Step 1: Differentiate the parabola
We start by differentiating the equation of the parabola with respect to \( x \).

Given:
\[ y^2 = 4ax \]

Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \]

Using the chain rule on the left side:
\[ 2y \frac{dy}{dx} = 4a \]

Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

### Step 2: Find the slope at the given point
Next, we need to find the slope of the tangent line at the point \( (at^2, 2at) \).

Substituting \( y = 2at \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(at^2, 2at)} = \frac{2a}{2at} = \frac{1}{t} \]

### Step 3: Write the equation of the tangent line
The equation of the tangent line can be written using the point-slope form:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (at^2, 2at) \) and \( m = \frac{1}{t} \).

Substituting these values:
\[ y - 2at = \frac{1}{t}(x - at^2) \]

Multiplying through by \( t \) to eliminate the fraction:
\[ t(y - 2at) = x - at^2 \]

Rearranging gives:
\[ x - ty + 2at^2 = 0 \]

### Step 4: Write the equation of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -t \]

Using the point-slope form again for the normal line:
\[ y - 2at = -t(x - at^2) \]

Multiplying through by \( t \):
\[ t(y - 2at) = -x + at^2 \]

Rearranging gives:
\[ ty + x - 2at^2 = 0 \]

### Final Equations
Thus, the equations of the tangent and normal to the parabola \( y^2 = 4ax \) at the point \( (at^2, 2at) \) are:
1. Tangent: \( x - ty + 2at^2 = 0 \)
2. Normal: \( ty + x - 2at^2 = 0 \)