Question

The value of p such that the line joining (0,3), (5, -2) is a tangent to the curve $y = \frac{p}{x+1}$ is

• A. 2
• B. 4
• C. 3
• D. 1

Hint:

At the point of tangency, both the $y$-values and the slopes of the line and the curve must be equal.

Answer 1

The Correct Option is B

Step 1: Concept
The slope of the line must equal the derivative of the curve at the point of tangency.

Step 2: Meaning
The slope of the line through (0,3) and (5,-2) is $m = \frac{-2-3}{5-0} = -1$. The derivative of $y = \frac{p}{x+1}$ is $\frac{dy}{dx} = -\frac{p}{(x+1)^{2}}$.

Step 3: Analysis
Equating the slopes: $-1 = -\frac{p}{(x+1)^{2}} \implies (x+1)^{2} = p$. The line equation is $y - 3 = -1(x - 0) \implies y = 3 - x$. Substituting $y$ into the curve equation: $3 - x = \frac{p}{x+1} \implies (3-x)(x+1) = p$.

Step 4: Conclusion
From $(x+1)^{2} = p$ and $(3-x)(x+1) = p$, we have $x+1 = 3-x \implies 2x = 2 \implies x = 1$. Substituting $x=1$ into $(x+1)^{2}=p$ gives $p = (1+1)^{2} = 4$. Final Answer: (B)