Question

The slope of a tangent drawn at the point \( P(\alpha, \beta) \) lying on the curve \( y = \frac{1}{2x - 5} \) is \( -2 \). If \( P \) lies in the fourth quadrant, then \( \alpha - \beta = \)

• A. $ 4 $
• B. $ 3 $
• C. $ 2 $
• D. $ 1 $

Hint:

To solve problems involving tangents to curves, use the derivative to find the slope and solve for the coordinates of the point. Ensure the point satisfies the given conditions (e.g., quadrant).

Answer 1

The Correct Option is B

Step 1: Find the derivative of the curve. 
The given curve is: $$ y = \frac{1}{2x - 5}. $$ Differentiate with respect to $ x $ using the chain rule: $$ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2x - 5} \right) = -\frac{1}{(2x - 5)^2} \cdot 2 = -\frac{2}{(2x - 5)^2}. $$ Step 2: Use the given slope. 
The slope of the tangent at $ P(\alpha, \beta) $ is given as $-2$. Thus: $$ -\frac{2}{(2\alpha - 5)^2} = -2. $$ Simplify: $$ \frac{2}{(2\alpha - 5)^2} = 2 \quad \Rightarrow \quad (2\alpha - 5)^2 = 1. $$ Take the square root: $$ 2\alpha - 5 = \pm 1. $$ Solve for $ \alpha $: 1. $ 2\alpha - 5 = 1 \quad \Rightarrow \quad 2\alpha = 6 \quad \Rightarrow \quad \alpha = 3 $, 2. $ 2\alpha - 5 = -1 \quad \Rightarrow \quad 2\alpha = 4 \quad \Rightarrow \quad \alpha = 2 $. 
Step 3: Determine $ \beta $ using the curve equation. 
Substitute $ \alpha = 3 $ and $ \alpha = 2 $ into the curve equation $ y = \frac{1}{2x - 5} $: 1. For $ \alpha = 3 $: $$ \beta = \frac{1}{2(3) - 5} = \frac{1}{6 - 5} = 1. $$ Thus, $ P(3, 1) $. 2. For $ \alpha = 2 $: $$ \beta = \frac{1}{2(2) - 5} = \frac{1}{4 - 5} = -1. $$ Thus, $ P(2, -1) $. 
Step 4: Identify the correct point in the fourth quadrant. 
The fourth quadrant requires $ \alpha>0 $ and $ \beta<0 $. Therefore, the correct point is $ P(2, -1) $. 
Step 5: Compute $ \alpha - \beta $. 
For $ P(2, -1) $: $$ \alpha - \beta = 2 - (-1) = 2 + 1 = 3. $$ 
Step 6: Final Answer. 
$$ \boxed{3} $$