Math · Precalculus · Worked example
Find the vertices and asymptotes of a hyperbola
Find the vertices, asymptotes and foci of the hyperbola 4x² − 9y² = 36.
Divide to get 1 on the right
Divide every term by 36.
Read a and b
The positive term is the x² term, so the branches open left and right. a² = 9 and b² = 4, so the vertices are (±3, 0).
Find the asymptotes
Far from the center the 1 is small compared with the squared terms, so x²/9 ≈ y²/4. That gives the lines y = ±(b/a)x.
Find the foci
For a hyperbola c² = a² + b² = 13, so the foci are (±√13, 0) ≈ (±3.61, 0), outside the vertices.
Result
Vertices (±3, 0), asymptotes y = ±(2/3)x, foci (±√13, 0) ≈ (±3.61, 0).
Your turn
Give the asymptotes and foci of y²/16 − x²/9 = 1.
Show the answer and explanation
Asymptotes y = ±(4/3)x; foci (0, ±5).
The positive term is the y² term, so the branches open up and down with a = 4 and b = 3. The asymptotes are y = ±(a/b)x = ±(4/3)x, and c² = 16 + 9 = 25.
Keep exploring
In Implicit curves & inequality regions, both branches appear. Swap the signs to −4x² + 9y² = 36 and the branches open up and down instead, with vertices (0, ±2).
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