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

Options

A. `(2,7)`
B. `(-2,7)`
C. `(-2,-7)`
D. `(2,-7)`

Answer

To solve the problem, we need to find the values of \( p \) and \( q \) given that the line \( y = 4x - 5 \) is a tangent to the curve defined by the equation \( y^2 = px^3 + q \) at the point \( (2, 3) \).

### Step-by-Step Solution:

1. **Identify the given equations**:
- The equation of the tangent line is \( y = 4x - 5 \).
- The equation of the curve is \( y^2 = px^3 + q \).

2. **Differentiate the curve**:
- Differentiate both sides of \( y^2 = px^3 + q \) with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(px^3 + q) \]
Using the chain rule on the left side, we get:
\[ 2y \frac{dy}{dx} = 3px^2 \]
- Rearranging gives:
\[ \frac{dy}{dx} = \frac{3px^2}{2y} \]

3. **Find the slope of the tangent line**:
- The slope of the tangent line \( y = 4x - 5 \) is \( 4 \).

4. **Evaluate the derivative at the point (2, 3)**:
- Substitute \( x = 2 \) and \( y = 3 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(2, 3)} = \frac{3p(2^2)}{2(3)} = \frac{3p \cdot 4}{6} = 2p \]
- Set this equal to the slope of the tangent line:
\[ 2p = 4 \]
- Solving for \( p \):
\[ p = 2 \]

5. **Substitute the point (2, 3) into the curve equation**:
- Substitute \( x = 2 \) and \( y = 3 \) into the curve equation \( y^2 = px^3 + q \):
\[ 3^2 = p(2^3) + q \]
This simplifies to:
\[ 9 = p \cdot 8 + q \]
- Substitute \( p = 2 \):
\[ 9 = 2 \cdot 8 + q \]
\[ 9 = 16 + q \]
- Solving for \( q \):
\[ q = 9 - 16 = -7 \]

6. **Final values**:
- We find \( p = 2 \) and \( q = -7 \).

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