Find the equation of normal at point (4, 3) for the hyperbola ` 3x^(2) - 4y^(2) = 14`.

Answer

To find the equation of the normal at the point (4, 3) for the hyperbola given by the equation \( 3x^2 - 4y^2 = 14 \), we will follow these steps:

### Step 1: Differentiate the hyperbola equation
We start with the equation of the hyperbola:
\[ 3x^2 - 4y^2 = 14 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(3x^2) - \frac{d}{dx}(4y^2) = 0 \]
This gives:
\[ 6x - 8y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to solve for \(\frac{dy}{dx}\):
\[ 8y \frac{dy}{dx} = 6x \]
\[ \frac{dy}{dx} = \frac{6x}{8y} = \frac{3x}{4y} \]

### Step 3: Calculate the slope of the tangent at the point (4, 3)
Now, we need to find the slope of the tangent at the point (4, 3):
\[ \frac{dy}{dx} \bigg|_{(4, 3)} = \frac{3(4)}{4(3)} = \frac{12}{12} = 1 \]
So, the slope of the tangent line at the point (4, 3) is \( m = 1 \).

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

### 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 \( (x_1, y_1) = (4, 3) \) and \( m = -1 \):
\[ y - 3 = -1(x - 4) \]
This simplifies to:
\[ y - 3 = -x + 4 \]
\[ y + x = 7 \]

### Final Equation
Thus, the equation of the normal at the point (4, 3) for the hyperbola \( 3x^2 - 4y^2 = 14 \) is:
\[ x + y = 7 \]