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

We are given $y = \frac{x - 3}{x - 4}$.
To find the slope of the normal, we first compute $\frac{dy}{dx}$, the derivative.
Use quotient rule: If $y = \frac{u}{v}$, then $\frac{dy}{dx} = \frac{v.du/dx - u.dv/dx}{v^2}$.
Here, $u = x - 3$, $v = x - 4$
$\frac{dy}{dx} = \frac{(x - 4)(1) - (x - 3)(1)}{(x - 4)^2} = \frac{x - 4 - x + 3}{(x - 4)^2} = \frac{-1}{(x - 4)^2}$
At $x = 6$, $(x - 4)^2 = 4$, so slope of tangent is $\frac{-1}{4}$
The slope of the normal is the negative reciprocal of the tangent’s slope.
So, slope of normal = $-1 / \left( \frac{-1}{4} \right) = 4$.
Wait — that is positive 4, but the answer marked is (C) $-4$.
Let’s double-check: $\text{slope of tangent} = -1/4$, so $\text{slope of normal} = 4$
So actually, correct answer is (A) 4, not (C) $-4$.
Final correction: Correct Answer: (A) 4