Question

The slope of the normal to the curve \( y = \frac{x - 3}{x - 4} \) at \( x = 6 \) is

• A. 4
• B. $-\dfrac{1}{4}$
• C. $-4$
• D. $\dfrac{1}{4}$

Hint:

To find the slope of a normal, take the negative reciprocal of the slope of the tangent at that point.

Answer 1

The Correct Option is C

Given $y = \frac{x - 3}{x - 4}$.
First, compute the derivative $\frac{dy}{dx}$ to find the slope of the tangent.
Using the quotient rule, where $y = \frac{u}{v}$, the derivative is $\frac{dy}{dx} = \frac{v.\frac{du}{dx} - u.\frac{dv}{dx}}{v^2}$.
Let $u = x - 3$ and $v = x - 4$.
Then $\frac{dy}{dx} = \frac{(x - 4)(1) - (x - 3)(1)}{(x - 4)^2} = \frac{x - 4 - x + 3}{(x - 4)^2} = \frac{-1}{(x - 4)^2}$.
At $x = 6$, $(x - 4)^2 = (6 - 4)^2 = 2^2 = 4$. Thus, the slope of the tangent at $x = 6$ is $\frac{-1}{4}$.
The slope of the normal is the negative reciprocal of the tangent's slope.
Therefore, the slope of the normal is $-1 / \left( \frac{-1}{4} \right) = 4$.
Initial calculation indicates the slope of the normal is 4. Re-verification confirms: slope of tangent = $-1/4$, so slope of normal = 4.
The correct answer is (A) 4.
Final Answer: Correct Answer: (A) 4