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: Understand the vertical tangent condition.
A vertical tangent occurs where $ dx/dy=0 $. Differentiating with respect to $ y $ avoids the $ dy/dx\to\infty $ difficulty.
Step 2: Differentiate $ y^3-3xy+2=0 $ with respect to y.
\[ 3y^2 - 3\!\left(x+y\frac{dx}{dy}\right) = 0 \implies \frac{dx}{dy} = \frac{y^2-x}{y} \]
Step 3: Set $ dx/dy=0 $.
\[ y^2-x=0 \implies x=y^2 \]
Step 4: Substitute $ x=y^2 $ into the curve.
\[ y^3-3y^3+2=0 \implies -2y^3+2=0 \implies y=1, \quad x=1 \]
Step 5: Verify the point lies on the curve.
$ 1-3+2=0 $. Confirmed.
Step 6: State the set of points with vertical tangent.
\[ \boxed{\{(1,1)\}} \]