Find the equations of the normals to the curve : `2x^(2)-y^(2)=14`, which are parallel to the line `x+3y=6.`

Answer

To find the equations of the normals to the curve \(2x^2 - y^2 = 14\) that are parallel to the line \(x + 3y = 6\), we will follow these steps:

### Step 1: Find the slope of the given line
The equation of the line can be rewritten in slope-intercept form \(y = mx + c\):
\[ x + 3y = 6 \implies 3y = -x + 6 \implies y = -\frac{1}{3}x + 2 \]
Thus, the slope \(m\) of the line is \(-\frac{1}{3}\).

### Step 2: Differentiate the curve to find the slope of the tangent
We differentiate the curve \(2x^2 - y^2 = 14\) implicitly with respect to \(x\):
\[ \frac{d}{dx}(2x^2) - \frac{d}{dx}(y^2) = 0 \]
This gives:
\[ 4x - 2y \frac{dy}{dx} = 0 \]
Rearranging this, we find:
\[ 2y \frac{dy}{dx} = 4x \implies \frac{dy}{dx} = \frac{2x}{y} \]
This represents the slope of the tangent line at any point on the curve.

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{y}{2x} \]

### Step 4: Set the slope of the normal equal to the slope of the given line
Since we want the normals to be parallel to the line \(x + 3y = 6\), we set the slope of the normal equal to \(-\frac{1}{3}\):
\[ -\frac{y}{2x} = -\frac{1}{3} \]
Cross-multiplying gives:
\[ 3y = 2x \implies y = \frac{2}{3}x \]

### Step 5: Substitute \(y\) back into the curve equation
Now we substitute \(y = \frac{2}{3}x\) into the curve equation \(2x^2 - y^2 = 14\):
\[ 2x^2 - \left(\frac{2}{3}x\right)^2 = 14 \]
Calculating \(y^2\):
\[ \left(\frac{2}{3}x\right)^2 = \frac{4}{9}x^2 \]
Substituting this back:
\[ 2x^2 - \frac{4}{9}x^2 = 14 \]
Finding a common denominator:
\[ \frac{18}{9}x^2 - \frac{4}{9}x^2 = 14 \implies \frac{14}{9}x^2 = 14 \]
Multiplying both sides by \(9\):
\[ 14x^2 = 126 \implies x^2 = 9 \implies x = \pm 3 \]

### Step 6: Find corresponding \(y\) values
Using \(y = \frac{2}{3}x\):
- For \(x = 3\):
\[ y = \frac{2}{3}(3) = 2 \implies (3, 2) \]
- For \(x = -3\):
\[ y = \frac{2}{3}(-3) = -2 \implies (-3, -2) \]

### Step 7: Write the equations of the normals
Now we have two points: \((3, 2)\) and \((-3, -2)\).

**Normal at (3, 2):**
Using the point-slope form:
\[ y - 2 = -\frac{1}{3}(x - 3) \]
Simplifying:
\[ y - 2 = -\frac{1}{3}x + 1 \implies 3y - 6 = -x + 3 \implies x + 3y = 9 \]

**Normal at (-3, -2):**
Using the point-slope form:
\[ y + 2 = -\frac{1}{3}(x + 3) \]
Simplifying:
\[ y + 2 = -\frac{1}{3}x - 1 \implies 3y + 6 = -x - 3 \implies x + 3y = -9 \]

### Final Equations
The equations of the normals are:
1. \(x + 3y = 9\)
2. \(x + 3y = -9\)