Problem

For \((x~-~2)^2/9~-~(y~+~1)^2/4~=~1\), give the hyperbola's center, \(a\), \(b\), opening direction, vertices, and \(\text{both}~\text{asymptote}~\text{equations}\).

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For a hyperbola, the positive squared term identifies the transverse axis, so it also controls the opening and vertex direction. Take square roots of the denominators to build the centered guide rectangle. Its transverse half-length locates the vertices, and its diagonals supply both asymptotes, with the rise-over-run ratio determined by the rectangle's orientation.

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The center is (2, -1); a = 2, b = 3; the hyperbola opens vertical; its vertices are (2, -3), (2, 1); and its asymptotes are y + 1 = (2/3)(x - 2) and y + 1 = -(2/3)(x - 2).

The center is (2, -1); a = 3, b = 2; the hyperbola opens horizontal; its vertices are (-1, -1), (5, -1); and its asymptotes are y + 1 = (3/2)(x - 2) and y + 1 = -(3/2)(x - 2).

The center is (2, -1); a = 3, b = 2; the hyperbola opens horizontal; its vertices are (-1, -1), (5, -1); and its asymptotes are y + 1 = (2/3)(x - 2) and y + 1 = -(2/3)(x - 2).

The center is (-2, 1); a = 3, b = 2; the hyperbola opens horizontal; its vertices are (-5, 1), (1, 1); and its asymptotes are y - 1 = (2/3)(x + 2) and y - 1 = -(2/3)(x + 2).

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For \((x~-~2)^2/9~-~(y~+~1)^2/4~=~1\), give the hyperbola's center, \(a\), \(b\), opening direction, vertices, and \(\text{both}~\text{asymptote}~\text{equations}\).

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