Question:
The line joining the points \((0,3)\) and \((5,-2)\) is a tangent to the curve
\[
y=\frac{C}{x+1}
\]
then \(C=\)
The line joining the points \((0,3)\) and \((5,-2)\) is a tangent to the curve
\[
y=\frac{C}{x+1}
\]
then \(C=\)
Show Hint
If a line is tangent to a curve, the resulting equation after substitution must have equal roots. Hence, use the discriminant condition:
\[
D=0.
\]
Updated On: Jun 22, 2026
- \(1\)
- \(-2\)
- \(4\)
- \(5\)
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The Correct Option is C
Solution and Explanation
Step 1: Find the equation of the line joining the given points.
The given points are \[ (0,3) \] and \[ (5,-2) \] Slope of the line is \[ m=\frac{-2-3}{5-0} \] \[ m=\frac{-5}{5}=-1 \] Using point-slope form through \((0,3)\), \[ y-3=-1(x-0) \] \[ y=-x+3 \]
Step 2: Use the tangency condition.
The curve is \[ y=\frac{C}{x+1} \] At the point of tangency, the line and the curve intersect at exactly one point.
So, \[ \frac{C}{x+1}=-x+3 \] Multiplying by \((x+1)\), \[ C=(-x+3)(x+1) \] Expand: \[ C=-x^2+2x+3 \] Rearranging, \[ x^2-2x+(C-3)=0 \] Since the line is tangent to the curve, this quadratic must have equal roots.
Therefore, \[ D=0 \]
Step 3: Apply discriminant condition.
For \[ x^2-2x+(C-3)=0, \] the discriminant is \[ (-2)^2-4(1)(C-3)=0 \] \[ 4-4C+12=0 \] \[ 16-4C=0 \] \[ 4C=16 \] \[ C=4 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{4} \]
The given points are \[ (0,3) \] and \[ (5,-2) \] Slope of the line is \[ m=\frac{-2-3}{5-0} \] \[ m=\frac{-5}{5}=-1 \] Using point-slope form through \((0,3)\), \[ y-3=-1(x-0) \] \[ y=-x+3 \]
Step 2: Use the tangency condition.
The curve is \[ y=\frac{C}{x+1} \] At the point of tangency, the line and the curve intersect at exactly one point.
So, \[ \frac{C}{x+1}=-x+3 \] Multiplying by \((x+1)\), \[ C=(-x+3)(x+1) \] Expand: \[ C=-x^2+2x+3 \] Rearranging, \[ x^2-2x+(C-3)=0 \] Since the line is tangent to the curve, this quadratic must have equal roots.
Therefore, \[ D=0 \]
Step 3: Apply discriminant condition.
For \[ x^2-2x+(C-3)=0, \] the discriminant is \[ (-2)^2-4(1)(C-3)=0 \] \[ 4-4C+12=0 \] \[ 16-4C=0 \] \[ 4C=16 \] \[ C=4 \]
Step 4: Final conclusion.
Therefore, \[ \boxed{4} \]
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