Question

\(V\) is the set of points on the curve \(y^3-3xy+2=0\) where the tangent is vertical, then \(V=\)

• A. \(\phi\)
• B. \(\{(1,0)\}\)
• C. \(\{(1,1)\}\)
• D. \(\{(0,0),(1,1)\}\)

Hint:

For vertical tangents: \[ \frac{dy}{dx}=\frac{N}{D} \] the denominator must be zero while the numerator remains non-zero.

Answer 1

The Correct Option is C

Step 1: Differentiate the curve implicitly.
Given: \[ y^3-3xy+2=0 \] Differentiating with respect to \(x\): \[ 3y^2\frac{dy}{dx}-3\left(x\frac{dy}{dx}+y\right)=0 \] \[ 3y^2\frac{dy}{dx}-3x\frac{dy}{dx}-3y=0 \] \[ (3y^2-3x)\frac{dy}{dx}=3y \] Therefore, \[ \frac{dy}{dx}=\frac{y}{y^2-x} \]

Step 2: Condition for vertical tangent.
A tangent is vertical when \[ \frac{dy}{dx} \] becomes infinite.
So, denominator must be zero: \[ y^2-x=0 \] Thus, \[ x=y^2 \] Also, numerator should not be zero simultaneously unless checked carefully.

Step 3: Substitute into the curve equation.
Substitute \[ x=y^2 \] into \[ y^3-3xy+2=0 \] \[ y^3-3(y^2)y+2=0 \] \[ y^3-3y^3+2=0 \] \[ -2y^3+2=0 \] \[ y^3=1 \] \[ y=1 \] Hence, \[ x=y^2=1 \] So, the point is \[ (1,1) \]

Step 4: Verify numerator is non-zero.
At \((1,1)\), \[ y=1\neq 0 \] Thus, \[ \frac{dy}{dx} \] is indeed infinite.

Step 5: Final conclusion.
Therefore, \[ \boxed{\{(1,1)\}} \]