Locate Vertices along the Positive-Term Axis

Supplied materials

• Paper or a browser scratch area; a calculator is optional unless rounding is requested.
• Supplied teaching case: Find the vertices of x²/4−y²/9=1. Candidate answers: (A) (0,−3) and (0,3) (B) (−2,0) and (2,0) (C) (−3,0) and (3,0) (D) (−4,0) and (4,0).
• Independent check inputs: Find the vertices of x²/9−y²/16=1. Candidate answers: (A) (0,−4) and (0,4) (B) (−3,0) and (3,0) (C) (−4,0) and (4,0) (D) (−9,0) and (9,0).

Learn the idea

Circle, parabola, ellipse and hyperbola equations encode geometric features through squared quantities and sign patterns. Standard forms locate centers, focal information, axes or vertices, with care to distinguish full lengths from half lengths. An origin-centered horizontal hyperbola has a positive x² term and a subtracted y² term. Setting y to zero locates the two vertices at plus and minus its horizontal semi-axis length.

Here is the method on the supplied teaching case. The positive x² term gives a horizontal transverse axis. Set y=0 to get x²/4=1, so x=±2.

The positive squared term decides the transverse-axis direction. The y denominator does not give these vertices; distinguish hyperbola subtraction from an ellipse’s sum and keep the square root of the x denominator.

1. Write or sketch the supplied teaching case, labeling its quantities and restrictions.
2. Apply the method shown in the worked solution, keeping each intermediate equality or justified inference.
3. Solve the independent check with its changed inputs before revealing feedback.
4. Complete the three practice cases; explain which mathematical rule rules out one distractor in each case.

Check your reasoning

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Take it further

Changed-input task: Find the vertices of x²/81−y²/100=1. Candidate answers: (A) (0,−10) and (0,10) (B) (−10,0) and (10,0) (C) (−81,0) and (81,0) (D) (−9,0) and (9,0). Show the mathematical reason, not only the selected answer. Expected result: (−9,0) and (9,0). The positive x² term gives a horizontal transverse axis. Set y=0 to get x²/81=1, so x=±9.

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