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

Options

A. p=2, q=7
B. p=-2, q=7
C. p=-2, q=-7
D. p=2, q=7

Answer

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

### Step 1: Substitute the point into the curve equation
We start by substituting the point \( (2, 3) \) into the curve equation \( y^2 = px^3 + q \).

\[ 3^2 = p(2^3) + q \]

Calculating the left side:

\[ 9 = p(8) + q \]

This simplifies to:

\[ 9 = 8p + q \quad \text{(Equation 1)} \]

### Step 2: Differentiate the curve equation
Next, we differentiate the curve equation \( y^2 = px^3 + q \) with respect to \( x \).

Using implicit differentiation:

\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(px^3 + q) \]

This gives:

\[ 2y \frac{dy}{dx} = 3px^2 \]

From this, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{3px^2}{2y} \]

### Step 3: Evaluate the derivative at the point (2, 3)
Now we evaluate \( \frac{dy}{dx} \) at the point \( (2, 3) \):

\[ \frac{dy}{dx} = \frac{3p(2^2)}{2(3)} = \frac{3p(4)}{6} = 2p \]

### Step 4: Compare the slopes
The slope of the tangent line \( y = 4x - 5 \) is \( 4 \). Therefore, we set the derivative equal to this slope:

\[ 2p = 4 \]

Solving for \( p \):

\[ p = \frac{4}{2} = 2 \]

### Step 5: Substitute \( p \) back into Equation 1
Now that we have \( p = 2 \), we substitute this value back into Equation 1 to find \( q \):

\[ 9 = 8(2) + q \]

This simplifies to:

\[ 9 = 16 + q \]

Solving for \( q \):

\[ q = 9 - 16 = -7 \]

### Final Values
Thus, we find:

\[ p = 2, \quad q = -7 \]

### Conclusion
The correct option is \( p = 2 \) and \( q = -7 \).

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