The equation of the normal to the ellipse `x^(2)/4+y^(2)/1=1` at (2, -1) is

Options

A. x-y-13=0
B. 5x-y+8=0
C. 3x+3y-3=0
D. 2x+y-3=0

Answer

To find the equation of the normal to the ellipse given by the equation \( \frac{x^2}{4} + \frac{y^2}{1} = 1 \) at the point \( (2, -1) \), we will follow these steps:

### Step 1: Differentiate the equation of the ellipse
The equation of the ellipse is:
\[ \frac{x^2}{4} + \frac{y^2}{1} = 1 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx} \left( \frac{x^2}{4} \right) + \frac{d}{dx} \left( \frac{y^2}{1} \right) = 0 \]
This gives:
\[ \frac{2x}{4} + 2y \frac{dy}{dx} = 0 \]
Simplifying this, we have:
\[ \frac{x}{2} + 2y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \( \frac{dy}{dx} \)
Rearranging the equation:
\[ 2y \frac{dy}{dx} = -\frac{x}{2} \]
Thus,
\[ \frac{dy}{dx} = -\frac{x}{4y} \]

### Step 3: Substitute the point \( (2, -1) \)
Now we substitute \( x = 2 \) and \( y = -1 \) into the derivative:
\[ \frac{dy}{dx} = -\frac{2}{4(-1)} = \frac{1}{2} \]
This value represents the slope of the tangent line at the point \( (2, -1) \).

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

### Step 5: Use the point-slope form to find the equation of the normal
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (2, -1) \) and \( m = -2 \):
\[ y - (-1) = -2(x - 2) \]
This simplifies to:
\[ y + 1 = -2x + 4 \]
Rearranging gives:
\[ 2x + y + 1 - 4 = 0 \]
Thus,
\[ 2x + y - 3 = 0 \]

### Final Answer
The equation of the normal to the ellipse at the point \( (2, -1) \) is:
\[ 2x + y - 3 = 0 \]