Find the equation of the normal at a point on the curve `x^(2)=4y`, which passes through the point (1,2). Also find the equation of the corresponding tangent.

Answer

To solve the problem of finding the equation of the normal at a point on the curve \( x^2 = 4y \) that passes through the point (1, 2), and also to find the equation of the corresponding tangent, we will follow these steps:

### Step 1: Identify the curve and point
The given curve is \( x^2 = 4y \). We can express \( y \) in terms of \( x \):
\[ y = \frac{x^2}{4} \]
We need to find a point \( (x_1, y_1) \) on this curve such that the normal at this point passes through (1, 2).

### Step 2: Find the derivative
To find the slope of the tangent line at any point on the curve, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left(\frac{x^2}{4}\right) = \frac{2x}{4} = \frac{x}{2} \]
The slope of the tangent line at the point \( (x_1, y_1) \) is:
\[ m_t = \frac{x_1}{2} \]
The slope of the normal line, which is perpendicular to the tangent, is given by:
\[ m_n = -\frac{1}{m_t} = -\frac{2}{x_1} \]

### Step 3: Write the equation of the normal
The equation of the normal line at the point \( (x_1, y_1) \) can be expressed using the point-slope form:
\[ y - y_1 = m_n (x - x_1) \]
Substituting \( y_1 = \frac{x_1^2}{4} \) and \( m_n = -\frac{2}{x_1} \):
\[ y - \frac{x_1^2}{4} = -\frac{2}{x_1}(x - x_1) \]
Simplifying this:
\[ y - \frac{x_1^2}{4} = -\frac{2}{x_1}x + 2 \]
\[ y = -\frac{2}{x_1}x + 2 + \frac{x_1^2}{4} \]

### Step 4: Substitute the point (1, 2)
Since the normal passes through the point (1, 2), we substitute \( x = 1 \) and \( y = 2 \):
\[ 2 = -\frac{2}{x_1}(1) + 2 + \frac{x_1^2}{4} \]
This simplifies to:
\[ 2 = -\frac{2}{x_1} + 2 + \frac{x_1^2}{4} \]
Subtracting 2 from both sides:
\[ 0 = -\frac{2}{x_1} + \frac{x_1^2}{4} \]
Multiplying through by \( 4x_1 \) to eliminate the fraction:
\[ 0 = -8 + x_1^3 \]
Thus, we have:
\[ x_1^3 = 8 \implies x_1 = 2 \]

### Step 5: Find \( y_1 \)
Now substituting \( x_1 = 2 \) back into the equation of the curve to find \( y_1 \):
\[ y_1 = \frac{2^2}{4} = 1 \]
So the point on the curve is \( (2, 1) \).

### Step 6: Write the normal equation
Now substituting \( x_1 = 2 \) and \( y_1 = 1 \) into the normal equation:
\[ y - 1 = -\frac{2}{2}(x - 2) \]
This simplifies to:
\[ y - 1 = -1(x - 2) \]
\[ y - 1 = -x + 2 \]
\[ y + x = 3 \]

### Step 7: Write the tangent equation
The slope of the tangent at \( (2, 1) \) is:
\[ m_t = \frac{2}{2} = 1 \]
Using the point-slope form for the tangent:
\[ y - 1 = 1(x - 2) \]
This simplifies to:
\[ y - 1 = x - 2 \]
\[ y - x = -1 \]

### Final Answers
- The equation of the normal is \( x + y = 3 \).
- The equation of the tangent is \( y - x = -1 \).