Question

The line joining the points \((0,3)\) and \((5,-2)\) is a tangent to the curve \[ y=\frac{C}{x+1} \] then \(C=\)

• A. \(1\)
• B. \(-2\)
• C. \(4\)
• D. \(5\)

Hint:

If a line is tangent to a curve, the resulting equation after substitution must have equal roots. Hence, use the discriminant condition: \[ D=0. \]

Answer 1

The Correct Option is C

Step 1: Find the equation of the line through $(0,3)$ and $(5,-2)$.
Slope $= \frac{-2 - 3}{5 - 0} = -1$. Equation: $y = -x + 3$.
Step 2: Find the slope of the tangent to $y = C/(x+1)$.
$\frac{dy}{dx} = -\frac{C}{(x+1)^2}$. For the line to be tangent, this slope must equal $-1$: $-\frac{C}{(x+1)^2} = -1$, so $(x+1)^2 = C$.
Step 3: Find the point of tangency on the curve.
At the tangent point, $(x+1)^2 = C$, so $x+1 = \sqrt{C}$ (taking positive root) and $y = \frac{C}{x+1} = \frac{C}{\sqrt{C}} = \sqrt{C}$. The point of tangency is $(\sqrt{C}-1,\, \sqrt{C})$.
Step 4: Impose that this point lies on the tangent line $y = -x+3$.
$\sqrt{C} = -(\sqrt{C}-1) + 3 = -\sqrt{C} + 4$. So $2\sqrt{C} = 4$, giving $\sqrt{C} = 2$, hence $C = 4$.
Step 5: Verify the answer.
With $C = 4$, curve is $y = 4/(x+1)$. Point of tangency: $(1, 2)$. Slope at $(1,2)$: $-4/(1+1)^2 = -1$. Tangent line: $y - 2 = -1(x-1)$, i.e., $y = -x+3$. Check $(0,3)$: $y=3$. Check $(5,-2)$: $y=-2$. Both lie on the line.
Step 6: State the final answer.
\[\boxed{C = 4}\]