Answer:
[tex]\begin{equation*} \frac{y^2}{9}-\frac{x^2}{16}=1 \end{equation*}[/tex]
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:
[tex]\frac{y^2}{a^2}-\frac{x^2}{b^2}=1[/tex]
From the graph:
[tex]\begin{gathered} (0,a)=(0,3) \\ (0,-a)=(0,-3) \\ \implies a=3 \end{gathered}[/tex]
Next, the equation of the asymptote of a hyperbola with a vertical transverse axis is:
[tex]y=\frac{a}{b}x[/tex]
Picking the points (0,0) and (4,3) on the line, find the slope of the asymptote:
[tex]\begin{gathered} Slope=\frac{3}{4} \\ \implies\frac{a}{b}=\frac{3}{4} \\ \implies b=4 \end{gathered}[/tex]
Therefore, the equation of the hyperbola is:
[tex]\begin{gathered} \frac{y^2}{3^2}-\frac{x^2}{4^2}=1 \\ \implies\frac{y^2}{9}-\frac{x^2}{16}=1 \end{gathered}[/tex]
