Question:
The tangent to the curve \(y^2-xy+9 = 0\) is vertical when
The tangent to the curve \(y^2-xy+9 = 0\) is vertical when
Show Hint
A vertical tangent means dy/dx is infinite, so the denominator of dy/dx is zero.
Updated On: Oct 1, 2026
- \(y = 0\)
- \(y = \pm \sqrt{3}\)
- \(y = \frac{1}{2}\)
- \(y = \pm 3\)
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The Correct Option is D
Solution and Explanation
Step 1: Differentiate implicitly:
\(2yy'-y-xy'=0\), so \(y'=\dfrac{y}{2y-x}\).
Step 2: Vertical tangent condition:
The slope is infinite when the denominator is zero: \(2y-x=0\), so \(x=2y\).
Step 3: Use the curve equation:
Substitute \(x=2y\): \(y^2-2y^2+9=0\), so \(y^2=9\) and \(y=\pm3\). Option D.
Step 4: Why the other options are wrong.
At \(y=0\) the curve equation gives \(9=0\), impossible. The values \(\pm\sqrt3\) and \(\tfrac12\) do not satisfy \(y^2=9\).
Final Answer:
The tangent is vertical when y = plus or minus 3. \[ \boxed{\text{(D) }y=\pm3} \]
\(2yy'-y-xy'=0\), so \(y'=\dfrac{y}{2y-x}\).
Step 2: Vertical tangent condition:
The slope is infinite when the denominator is zero: \(2y-x=0\), so \(x=2y\).
Step 3: Use the curve equation:
Substitute \(x=2y\): \(y^2-2y^2+9=0\), so \(y^2=9\) and \(y=\pm3\). Option D.
Step 4: Why the other options are wrong.
At \(y=0\) the curve equation gives \(9=0\), impossible. The values \(\pm\sqrt3\) and \(\tfrac12\) do not satisfy \(y^2=9\).
Final Answer:
The tangent is vertical when y = plus or minus 3. \[ \boxed{\text{(D) }y=\pm3} \]
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