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Hey Mahfia, please help! Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
y = ± [tex]\frac{5}{6} x[/tex]

Respuesta :

Nayefx

Answer:

[tex] \huge\boxed{ \red{ \boxed{ \tt{ \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1}}}}[/tex]

Step-by-step explanation:

to understand this

you need to know about:

  • conic sections
  • PEMDAS

tips and formulas:

  • [tex] \sf hyperbola \:equation : \\ \sf \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1[/tex]
  • vertices of hyperbola:(±a,0) and (0,±b) if reversed
  • [tex] \sf \: asymptotes : \\ y = \pm\frac{b}{a} x[/tex]

given:

  • vertices: (0,±10)
  • the hyperbola equation is inversed since the vertices is (0,±10)
  • asymptotes:[tex]\pm \frac{5}{6}x[/tex]

let's solve:

  • the asymptotes are in simplest and we know b is ±10

according to the question

  1. [tex]y = \sf \frac{5 \times 2}{6 \times 2} x \\ y = \frac{10}{12} x[/tex]

therefore we got

  • a=12
  • b=10

note: the equation will be inversed

let's create the equation:

  1. [tex] \sf substitute \: the \: value \: of \: a \: and \: b : \\ \sf \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1[/tex]
Ver imagen Nayefx

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