Equation of normal to the curve y = x (2 - x) at the point (2, 0) is

Options

A. x-2y=2
B. 2x+y=4
C. x-2y+2=0
D. None of these

Answer

To find the equation of the normal to the curve \( y = x(2 - x) \) at the point \( (2, 0) \), we will follow these steps:

### Step 1: Find the derivative of the curve
The curve is given by:
\[ y = x(2 - x) = 2x - x^2 \]
To find the slope of the tangent line at any point on the curve, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2 - 2x \]

### Step 2: Evaluate the derivative at the point (2, 0)
Next, we substitute \( x = 2 \) into the derivative to find the slope of the tangent line at the point \( (2, 0) \):
\[ \frac{dy}{dx} \bigg|_{x=2} = 2 - 2(2) = 2 - 4 = -2 \]
So, the slope of the tangent line at the point \( (2, 0) \) is \( -2 \).

### Step 3: Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, if the slope of the tangent line is \( -2 \), the slope of the normal line \( m \) is:
\[ m = -\frac{1}{\text{slope of tangent}} = -\frac{1}{-2} = \frac{1}{2} \]

### Step 4: Use the point-slope form to find the equation of the normal
We can use the point-slope form of the equation of a line, which is given by:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (2, 0) \) and \( m = \frac{1}{2} \):
\[ y - 0 = \frac{1}{2}(x - 2) \]
This simplifies to:
\[ y = \frac{1}{2}x - 1 \]

### Step 5: Rearranging to standard form
To express this in standard form, we can rearrange the equation:
\[ \frac{1}{2}x - y - 1 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ x - 2y - 2 = 0 \]
Thus, the equation of the normal line is:
\[ x - 2y = 2 \]

### Final Answer
The equation of the normal to the curve \( y = x(2 - x) \) at the point \( (2, 0) \) is:
\[ x - 2y = 2 \]

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