Problem

Complete the squares and normalize \(x^2~-~y^2~-~4x~-~6y~=~6\). Give the hyperbola's standard form, center, \(a\), \(b\), opening direction, and vertices.

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Preserve the subtraction on an entire variable group while completing both squares; distributing that outer negative is the delicate step. After simplifying, scale the equation to a positive one on the right. The positive squared term determines the transverse and opening direction, its denominator is the squared vertex distance, and the center comes from the opposite signs inside the binomials.

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

The standard form is (x - 2)²/6 - (y + 3)²/6 = 1; the center is (2, -3); a = sqrt(6) and b = sqrt(6); the hyperbola opens horizontal; and its vertices are (2 - sqrt(6), -3), (2 + sqrt(6), -3).

The standard form is (y + 3)²/1 - (x - 2)²/1 = 1; the center is (2, -3); a = 1 and b = 1; the hyperbola opens vertical; and its vertices are (2, -4), (2, -2).

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

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Complete the squares and normalize \(x^2~-~y^2~-~4x~-~6y~=~6\). Give the hyperbola's standard form, center, \(a\), \(b\), opening direction, and vertices.

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