Question

The tangent to the curve \(y^2-xy+9 = 0\) is vertical when

• A. \(y = 0\)
• B. \(y = \pm \sqrt{3}\)
• C. \(y = \frac{1}{2}\)
• D. \(y = \pm 3\)

Hint:

A vertical tangent means dy/dx is infinite, so the denominator of dy/dx is zero.

Answer 1

The Correct Option is D

Step 1: Treat x as a function of y:
Write the curve as $x=\dfrac{y^2+9}{y}=y+\dfrac9y$.

Step 2: Vertical tangent means dx/dy = 0:
$\dfrac{dx}{dy}=1-\dfrac{9}{y^2}=0$, so $y^2=9$ and $y=\pm3$.

Step 3: Pick:
Option D.

Final Answer:
Setting dx/dy = 0 gives y squared equal to 9. \[ \boxed{\text{(D) }y=\pm3} \]