Question

1. $\frac{(y - 2)^2}{4}-\frac{(x + 3)^2}{9}=1$

part i: what kind of conic section is this? (2 points)
part ii: what are the coordinates of its center? (2 points)
part iii: what are the coordinates of the two vertices? (2 points)
part iv: what are the coordinates of the foci? (2 points)
part v: what are the equations of the asymptotes? (2 points)

Explanation:

Step1: Identify conic - section type

The equation is of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, which is a hyperbola with a vertical transverse axis.

Step2: Find the center

For a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, the center is $(h,k)$. From the given equation $\frac{(y - 2)^2}{4}-\frac{(x+3)^2}{9}=1$, we have $h=-3$ and $k = 2$. So the center is $(-3,2)$.

Step3: Find the vertices

For a hyperbola $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, the vertices are $(h,k\pm a)$. Here $a^2 = 4$, so $a = 2$. The vertices are $(-3,2\pm2)$, i.e., $(-3,0)$ and $(-3,4)$.

Step4: Find the foci

First, find $c$ using $c^2=a^2 + b^2$. Here $a^2 = 4$ and $b^2=9$, so $c^2=4 + 9=13$, $c=\sqrt{13}$. The foci are $(h,k\pm c)$, i.e., $(-3,2\pm\sqrt{13})$.

Step5: Find the asymptotes

The equations of the asymptotes for a hyperbola $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$ are $y - k=\pm\frac{a}{b}(x - h)$. Here $a = 2$ and $b = 3$, so the asymptotes are $y-2=\pm\frac{2}{3}(x + 3)$.

Answer:

Part I: Hyperbola
Part II: $(-3,2)$
Part III: $(-3,0),(-3,4)$
Part IV: $(-3,2-\sqrt{13}),(-3,2+\sqrt{13})$
Part V: $y-2=\frac{2}{3}(x + 3)$ and $y-2=-\frac{2}{3}(x + 3)$