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

Options

A. meets the curve again the third quadrant
B. meets the curve again in the fourth quadrant
C. does not meet the curve again
D. meets the curve again in the second 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:
\[ x^2 + 2xy - 3y^2 = 0 \]
We differentiate 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 \(3y^2\), we get:
\[ 2x + 2\left(x\frac{dy}{dx} + y\right) - 6y\frac{dy}{dx} = 0 \]

### Step 2: Rearranging the differentiated equation
Rearranging gives:
\[ 2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0 \]
Combining the terms involving \(\frac{dy}{dx}\):
\[ 2x + 2y + (2x - 6y)\frac{dy}{dx} = 0 \]

### Step 3: Solve for \(\frac{dy}{dx}\)
Now, we isolate \(\frac{dy}{dx}\):
\[ (2x - 6y)\frac{dy}{dx} = - (2x + 2y) \]
Thus,
\[ \frac{dy}{dx} = -\frac{2x + 2y}{2x - 6y} \]

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

### Step 5: 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:
\[ m_{\text{normal}} = -\frac{1}{1} = -1 \]

### Step 6: Write the equation of the normal
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \((x_1, y_1) = (1, 1)\) and \(m = -1\):
\[ y - 1 = -1(x - 1) \]
This simplifies to:
\[ y - 1 = -x + 1 \]
or
\[ y + x - 2 = 0 \]

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