Find the eqaution of the normal to `x^2+y^2=5` at the point (2,1)

Options

A. x+2y=0
B. x-2y=0
C. y-2x=0
D. y+2x=0

Answer

To find the equation of the normal to the curve \(x^2 + y^2 = 5\) at the point \((2, 1)\), we will follow these steps:

### Step 1: Differentiate the equation of the curve
We start with the equation of the curve:
\[ x^2 + y^2 = 5 \]
We will differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(5) \]
Using the chain rule for \(y^2\):
\[ 2x + 2y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation gives:
\[ 2y \frac{dy}{dx} = -2x \]
Dividing both sides by \(2y\):
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Step 3: Find the slope of the tangent at the point (2, 1)
Now we will substitute the coordinates of the point \((2, 1)\) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(2, 1)} = -\frac{2}{1} = -2 \]
Thus, the slope of the tangent line at the point \((2, 1)\) is \(-2\).

### Step 4: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent. Therefore:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{-2} = \frac{1}{2} \]

### Step 5: Use the point-slope form to find the equation of the normal
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1) = (2, 1)\) and \(m = \frac{1}{2}\):
\[ y - 1 = \frac{1}{2}(x - 2) \]

### Step 6: Simplify the equation
Multiplying both sides by 2 to eliminate the fraction:
\[ 2(y - 1) = x - 2 \]
Expanding gives:
\[ 2y - 2 = x - 2 \]
Rearranging this gives:
\[ x - 2y = 0 \]

### Final Equation
Thus, the equation of the normal to the curve at the point \((2, 1)\) is:
\[ x - 2y = 0 \]

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