Recover a vertical hyperbola from vertices and asymptotes

Use the vertical form $(y-k)^2/b^2-(x-h)^2/a^2=1$. The centre is $(-1,2)$, so $h=-1,k=2$. The vertices $(-1,5),(-1,-1)$ are each $3$ units vertically from the centre, so $b=3$. This is the Teacher copy’s alternative-method note: the given vertex ordinates fix $b$.

The given lines $y=\tfrac34x+\tfrac{11}{4}$ and $y=-\tfrac34x+\tfrac54$ rewrite as $y=2\pm\tfrac34(x+1)$. Compare with $y=k\pm(b/a)(x-h)$: $b/a=3/4$. Together with $b=3$, this gives $a=4$. The ratio alone does not fix the sizes of $a$ and $b$; the vertex distance supplies the missing scale.

The equation is $\dfrac{(y-2)^2}{3^2}-\dfrac{(x+1)^2}{4^2}=1$. It gives vertices $(-1,2\pm3)$ and asymptote slopes $\pm3/4$ through $(-1,2)$, matching all the diagram labels.