The normal to the curve `x^(2)+2xy-3y^(2)=0,` at (1,1)

Options

A. does not meet the curve again.
B. meets the curve again in the second quadrant.
C. meets the curve again in the third quadrant.
D. meets the curve again in the fourth quadrant.

Answer

To find the normal to the curve \( x^2 + 2xy - 3y^2 = 0 \) at the point \( (1, 1) \), we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation of the curve:
\[ x^2 + 2xy - 3y^2 = 0 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) - \frac{d}{dx}(3y^2) = 0 \]
Using the product rule for \( 2xy \) and the chain rule for \( 3y^2 \):
\[ 2x + 2(y + x\frac{dy}{dx}) - 6y\frac{dy}{dx} = 0 \]
This simplifies to:
\[ 2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0 \]

### Step 2: Rearrange to isolate \( \frac{dy}{dx} \)
Rearranging the equation gives:
\[ 2x + 2y = (6y - 2x)\frac{dy}{dx} \]
Thus, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = \frac{2x + 2y}{6y - 2x} \]

### Step 3: Substitute the point \( (1, 1) \)
Now we substitute \( x = 1 \) and \( y = 1 \) into the derivative:
\[ \frac{dy}{dx} = \frac{2(1) + 2(1)}{6(1) - 2(1)} = \frac{2 + 2}{6 - 2} = \frac{4}{4} = 1 \]

### Step 4: Find the slope of the normal
The slope of the tangent line at the point \( (1, 1) \) is \( 1 \). Therefore, the slope of the normal line, which is perpendicular to the tangent, is:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -1 \]

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

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

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