If y = 4x-5 is a tangent to the curve `y^2 = ax^3 + b` at (2,3), then

Options

A. a=2,b=-7
B. a=-2,b=7
C. a=-2,b=-7
D. a=2,b=7

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the line \( y = 4x - 5 \) is a tangent to the curve \( y^2 = ax^3 + b \) at the point \( (2, 3) \).

### Step-by-Step Solution:

1. **Substitute the point into the curve equation**:
Since the point \( (2, 3) \) lies on the curve \( y^2 = ax^3 + b \), we substitute \( x = 2 \) and \( y = 3 \) into the equation:
\[ 3^2 = a(2^3) + b \]
Simplifying this gives:
\[ 9 = 8a + b \quad \text{(Equation 1)} \]

2. **Find the slope of the tangent line**:
The slope of the tangent line given is \( m = 4 \). We will find the slope of the curve at the point \( (2, 3) \) using implicit differentiation. The curve is given by:
\[ y^2 = ax^3 + b \]
Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 3ax^2 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{3ax^2}{2y} \]

3. **Substitute the point into the derivative**:
Now we substitute \( x = 2 \) and \( y = 3 \) into the derivative to find the slope at that point:
\[ \frac{dy}{dx} = \frac{3a(2^2)}{2(3)} = \frac{12a}{6} = 2a \]

4. **Set the slopes equal**:
Since the slope of the tangent line must equal the slope of the curve at the point of tangency, we set:
\[ 2a = 4 \]
Solving for \( a \) gives:
\[ a = 2 \]

5. **Substitute \( a \) back into Equation 1**:
Now that we have \( a = 2 \), we substitute it back into Equation 1:
\[ 9 = 8(2) + b \]
Simplifying this gives:
\[ 9 = 16 + b \implies b = 9 - 16 = -7 \]

6. **Final values**:
We have found:
\[ a = 2 \quad \text{and} \quad b = -7 \]

### Conclusion:
The values of \( a \) and \( b \) are:
\[ \boxed{a = 2, b = -7} \]