The slope of normal to the curve `y = (x -1)/(x - 2),x ne 2, at x = 10` is:

Options

A. 10
B. 64
C. 100
D. 8

Answer

To find the slope of the normal to the curve \( y = \frac{x - 1}{x - 2} \) at \( x = 10 \), we will follow these steps:

### Step 1: Find the derivative \( \frac{dy}{dx} \)
We start by differentiating the function using the quotient rule. The quotient rule states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).

Here, \( u = x - 1 \) and \( v = x - 2 \).

- Calculate \( \frac{du}{dx} = 1 \)
- Calculate \( \frac{dv}{dx} = 1 \)

Now applying the quotient rule:
\[ \frac{dy}{dx} = \frac{(x - 2)(1) - (x - 1)(1)}{(x - 2)^2} \]

### Step 2: Simplify the derivative
Now we simplify the expression:
\[ \frac{dy}{dx} = \frac{x - 2 - (x - 1)}{(x - 2)^2} = \frac{x - 2 - x + 1}{(x - 2)^2} = \frac{-1}{(x - 2)^2} \]

### Step 3: Evaluate the derivative at \( x = 10 \)
Now we need to find the slope of the tangent line at \( x = 10 \):
\[ \frac{dy}{dx} \bigg|_{x=10} = \frac{-1}{(10 - 2)^2} = \frac{-1}{8^2} = \frac{-1}{64} \]

### Step 4: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore:
\[ \text{slope of normal} = -\frac{1}{\frac{-1}{64}} = 64 \]

### Final Answer
Thus, the slope of the normal to the curve at \( x = 10 \) is \( 64 \).

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