Question

If the line \( 6x - y - 4 = 0 \) touches the curve \( y^2 = ax^3 + b \) at the point (1, 2), then \( a + b = \)

• A. 8
• B. -4
• C. 4
• D. 12

Hint:

For tangency conditions, differentiate the curve equation and apply the point of tangency to find the values of constants.

Answer 1

The Correct Option is C

Step 1: Differentiate the curve equation.
We are given the curve equation \( y^2 = ax^3 + b \). Differentiate both sides with respect to \( x \): \[ 2y \frac{dy}{dx} = 3ax^2 \]
Step 2: Find the slope at the point.
At the point \( (1, 2) \), we substitute \( x = 1 \) and \( y = 2 \) into the equation to find the slope \( \frac{dy}{dx} \).
Step 3: Use the point of tangency condition.
Using the point \( (1, 2) \) and the condition that the line is tangent to the curve, we solve for \( a \) and \( b \). We find that \( a + b = 4 \), corresponding to option (C).