Answer:
The standard form of this hyperbola is [tex]\frac{(x-3)^{2}}{5}-\frac{(y+1)^{2}}{9} = 1[/tex].
Step-by-step explanation:
From Analytical Geometry, the standard form of the hyperbola is defined by the following expression:
[tex]\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}} = 1[/tex] (1)
Where:
[tex]x[/tex] - Independent variable, dimensionless.
[tex]y[/tex] - Dependent variable, dimensionless.
[tex]a, b[/tex] - Semiaxis lengths, dimensionless.
[tex]h, k[/tex] - Coordinates of the center, dimensionless.
Let the hyperbola be defined by [tex]9\cdot x^{2}-5\cdot y^{2}-54\cdot x -10\cdot y +31=0[/tex], we proceed to derive the standard form by algebraic means:
1) [tex]9\cdot x^{2}-5\cdot y^{2}-54\cdot x -10\cdot y +31=0[/tex] Given
2) [tex](9\cdot x^{2}-54\cdot x)+(-1) \cdot (5\cdot y^{2}+10\cdot y)+31 = 0[/tex] Commutative and associative properties/[tex](-1)\cdot a = -a[/tex]
3) [tex](9\cdot x^{2}-54\cdot x + 81)+(-81)+(-1)\cdot (5\cdot y^{2}+10\cdot y +5)+(-1)\cdot (-5)+31 = 0[/tex] Modulative, distributive and associative properties/[tex](-1)\cdot a = -a[/tex]/Existence of the additive inverse
4) [tex](3\cdot x-9)^{2}+(-1)\cdot (\sqrt{5}\cdot y +\sqrt{5})^{2}+(-81)+5+31 = 0[/tex] Perfect trinomial square/[tex](-a)\cdot (-b) = a\cdot b[/tex]/Commutative property
5) [tex]9\cdot (x-3)^{2}+(-1)\cdot (5) \cdot (y+1)^{2}-45 = 0[/tex] Distributive property/[tex](a\cdot b)^{c} = a^{c}\cdot b^{c}[/tex]/Definition of addition and subtraction.
6) [tex]9\cdot (x-3)^{2}-5\cdot (y+1)^{2}= 45[/tex] [tex](-1)\cdot a = -a[/tex]/Compatibility with addition/Existence of additive inverse/Modulative property.
7) [tex]\frac{(x-3)^{2}}{5}-\frac{(y+1)^{2}}{9} = 1[/tex] Compatibility with multiplication/Definition of division/Distributive property/Result.
The standard form of this hyperbola is [tex]\frac{(x-3)^{2}}{5}-\frac{(y+1)^{2}}{9} = 1[/tex].
