The normal to the curve `x^(2)=4y` passing `(1, 2)` is :

Options

A. `x+y=3`
B. `x-y=3`
C. `x+y=1`
D. `x-y=1`.

Answer

To find the normal to the curve \( x^2 = 4y \) passing through the point \( (1, 2) \), we will follow these steps:

### Step 1: Differentiate the curve
The given curve is \( x^2 = 4y \). We differentiate both sides with respect to \( x \):

\[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4y) \]

This gives us:

\[ 2x = 4 \frac{dy}{dx} \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation from Step 1, we find:

\[ \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2} \]

### Step 3: Find the slope of the tangent at the point
We need to find the slope of the tangent line at the point where the normal passes through \( (1, 2) \). We first need to find the corresponding \( y \) value on the curve when \( x = 1 \):

\[ x^2 = 4y \implies 1^2 = 4y \implies 1 = 4y \implies y = \frac{1}{4} \]

This means the point \( (1, 2) \) is not on the curve. We need to find a point on the curve where the normal line passes through \( (1, 2) \).

### Step 4: Set up the normal line equation
Let the point on the curve be \( (x_1, y_1) \). From the curve equation:

\[ y_1 = \frac{x_1^2}{4} \]

The slope of the tangent at this point is:

\[ m_1 = \frac{x_1}{2} \]

The slope of the normal line \( m_2 \) is given by:

\[ m_2 = -\frac{1}{m_1} = -\frac{2}{x_1} \]

### Step 5: Use the point-slope form of the line
Using the point-slope form of the line, we can write the equation of the normal line passing through \( (x_1, y_1) \):

\[ y - y_1 = m_2(x - x_1) \]

Substituting \( y_1 = \frac{x_1^2}{4} \) and \( m_2 = -\frac{2}{x_1} \):

\[ y - \frac{x_1^2}{4} = -\frac{2}{x_1}(x - x_1) \]

### Step 6: Substitute the point (1, 2) into the normal line equation
Now we substitute \( (1, 2) \) into the equation:

\[ 2 - \frac{x_1^2}{4} = -\frac{2}{x_1}(1 - x_1) \]

### Step 7: Simplify and solve for \( x_1 \)
Rearranging gives:

\[ 2 - \frac{x_1^2}{4} = -\frac{2}{x_1} + 2 \]

This simplifies to:

\[ -\frac{x_1^2}{4} = -\frac{2}{x_1} \]

Multiplying through by \( -4x_1 \) gives:

\[ x_1^3 = 8 \implies x_1 = 2 \]

### Step 8: Find \( y_1 \)
Substituting \( x_1 = 2 \) back into the curve equation:

\[ y_1 = \frac{2^2}{4} = 1 \]

### Step 9: Write the normal line equation
Now we have the point \( (2, 1) \) on the curve. The slope of the normal at this point is:

\[ m_2 = -\frac{2}{2} = -1 \]

Using the point-slope form again:

\[ y - 1 = -1(x - 2) \]

This simplifies to:

\[ y - 1 = -x + 2 \implies y + x = 3 \]

### Final Answer
Thus, the equation of the normal to the curve \( x^2 = 4y \) passing through the point \( (1, 2) \) is:

\[ x + y = 3 \]