Question

Hey Mahfia, please help! Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at

Answer 1

Answer:

$\huge\boxed{ \red{ \boxed{ \tt{ \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1}}}}$

Step-by-step explanation:

to understand this

you need to know about:

• conic sections
• PEMDAS

tips and formulas:

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

given:

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

let's solve:

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

according to the question

1. $y = \sf \frac{5 \times 2}{6 \times 2} x \\ y = \frac{10}{12} x$

therefore we got

• a=12
• b=10

note: the equation will be inversed

let's create the equation:

1. $\sf substitute \: the \: value \: of \: a \: and \: b : \\ \sf \frac{ {y}^{2} }{ {10}^{2} } - \frac{ {x}^{2} }{ {12}^{2} } = 1$