Question

I need to come up with the standard equation for a picture of a graphed hyperbola, picture included

Answer 1

`\begin{equation*} (y^2)/(9)-(x^2)/(16)=1 \end{equation*}`

Explanation:

The hyperbola has a vertical transverse axis, and is centred at the origin, (0,0), therefore, the equation of the hyperbola will be an equation of the form:

`(y^2)/(a^2)-(x^2)/(b^2)=1`

From the graph:

`\begin{gathered} (0,a)=(0,3) \\ (0,-a)=(0,-3) \\ \implies a=3 \end{gathered}`

Next, the equation of the asymptote of a hyperbola with a vertical transverse axis is:

`y=(a)/(b)x`

Picking the points (0,0) and (4,3) on the line, find the slope of the asymptote:

`\begin{gathered} Slope=(3)/(4) \\ \implies(a)/(b)=(3)/(4) \\ \implies b=4 \end{gathered}`

Therefore, the equation of the hyperbola is:

`\begin{gathered} (y^2)/(3^2)-(x^2)/(4^2)=1 \\ \implies(y^2)/(9)-(x^2)/(16)=1 \end{gathered}`