Question

determine the center, vertices, and foci for the hyperbola.

\\(\frac{y^{2}}{36}-\frac{(x + 2)^{2}}{20}=1\\)

enter your answers as points \\((a,b)\\). enter the vertices and foci in increasing order of the \\(x\\)-coordinate.

the center is \\(\square\\).

the vertices are \\(\square\\) and \\(\square\\).

the foci are \\(\square\\) and \\(\square\\).

Explanation:

Step1: Identify the standard - form and center

The standard form of a hyperbola with a vertical transverse axis is $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, and the center is $(h,k)$. For the hyperbola $\frac{y^{2}}{36}-\frac{(x + 2)^{2}}{20}=1$, we have $h=-2,k = 0,a^{2}=36$, so $a = 6$, and $b^{2}=20$. The center is $(-2,0)$.

Step2: Find the vertices

The vertices of a hyperbola with a vertical transverse axis are given by $(h,k\pm a)$. Substituting $h=-2,k = 0,a = 6$, we get the vertices $(-2, - 6)$ and $(-2,6)$.

Step3: Calculate $c$ and find the foci

We know that $c^{2}=a^{2}+b^{2}$. Since $a^{2}=36$ and $b^{2}=20$, then $c^{2}=36 + 20=56$, so $c=\sqrt{56}=2\sqrt{14}$. The foci of a hyperbola with a vertical transverse axis are $(h,k\pm c)$. Substituting $h=-2,k = 0,c = 2\sqrt{14}$, we get the foci $(-2,-2\sqrt{14})$ and $(-2,2\sqrt{14})$.

Answer:

The center is $(-2,0)$.
The vertices are $(-2,-6)$ and $(-2,6)$.
The foci are $(-2,-2\sqrt{14})$ and $(-2,2\sqrt{14})$.