Question

If

\[ A = \{ P(\alpha, \beta) \mid \text{the tangent drawn at P to the curve } y^3 - 3xy + 2 = 0 \text{ is a horizontal line} \} \]

and

\[ B = \{ Q(a, b) \mid \text{the tangent drawn at Q to the curve } y^3 - 3xy + 2 = 0 \text{ is a vertical line} \} \]

then \( n(A) + n(B) = \)

• A. \( 12 \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( 4 \)

Hint:

For horizontal tangents, set \( \frac{dy}{dx} = 0 \). For vertical tangents, set the denominator of \( \frac{dy}{dx} \) to zero and solve for points.

Answer 1

The Correct Option is B

Step 1: Find conditions for horizontal and vertical tangents 
The given curve equation: \[ y^3 - 3xy + 2 = 0. \] Differentiate implicitly: \[ 3y^2 \frac{dy}{dx} - (3x \frac{dy}{dx} + 3y) = 0. \] \[ (3y^2 - 3x) \frac{dy}{dx} = 3y. \] \[ \frac{dy}{dx} = \frac{3y}{3y^2 - 3x}. \] 

Step 2: Condition for horizontal tangents 
For a horizontal tangent: \[ \frac{dy}{dx} = 0. \] \[ \frac{3y}{3y^2 - 3x} = 0. \] \[ 3y = 0 \Rightarrow y = 0. \] Substituting \( y = 0 \) in the curve equation: \[ 0^3 - 3x(0) + 2 = 0. \] \[ 2 = 0, \quad \text{(contradiction).} \] Thus, there are no points with horizontal tangents. \[ n(A) = 0. \]

 Step 3: Condition for vertical tangents 
For a vertical tangent, denominator of \( \frac{dy}{dx} \) should be zero: \[ 3y^2 - 3x = 0. \] \[ y^2 = x. \] Substituting \( x = y^2 \) in the curve equation: \[ y^3 - 3(y^2)y + 2 = 0. \] \[ y^3 - 3y^3 + 2 = 0. \] \[ -2y^3 + 2 = 0. \] \[ y^3 = 1. \] \[ y = 1. \] Substituting \( y = 1 \) in \( x = y^2 \): \[ x = 1^2 = 1. \] So, \( (1,1) \) is the only point where the tangent is vertical. \[ n(B) = 1. \] 

Step 4: Conclusion 
\[ n(A) + n(B) = 0 + 1 = 1. \] Thus, the correct answer is: \[ \mathbf{1}. \]