Question

for question 6, answer the questions about the conic section. (5 points)

1. $\frac{(x - 4)^2}{36}-\frac{(y - 2)^2}{9}=1$

center:
vertices:
foci:
eccentricity:
asymptotes:
length of transverse axis:
length of conjugate axis:

Explanation:

Step1: Identify the form of hyperbola

The equation $\frac{(x - 4)^2}{36}-\frac{(y - 2)^2}{9}=1$ is of the form $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, where $(h,k)$ is the center, $a^2 = 36$, $b^2=9$. So $a = 6$, $b = 3$.

Step2: Find the center

For the hyperbola $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$, the center is $(h,k)$. Here $h = 4$ and $k = 2$, so the center is $(4,2)$.

Step3: Find the vertices

The vertices of the hyperbola $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$ are $(h\pm a,k)$. Substituting $h = 4$, $a = 6$, $k = 2$, we get $(4\pm6,2)$, i.e., $(10,2)$ and $(- 2,2)$.

Step4: Find the foci

First, find $c$ using the formula $c^2=a^2 + b^2$. Since $a^2 = 36$ and $b^2=9$, then $c^2=36 + 9=45$, so $c=\sqrt{45}=3\sqrt{5}$. The foci are $(h\pm c,k)$, i.e., $(4\pm3\sqrt{5},2)$.

Step5: Find the eccentricity

The eccentricity $e$ of a hyperbola is given by $e=\frac{c}{a}$. Since $c = 3\sqrt{5}$ and $a = 6$, then $e=\frac{3\sqrt{5}}{6}=\frac{\sqrt{5}}{2}$.

Step6: Find the asymptotes

The equations of the asymptotes of the hyperbola $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$ are $y - k=\pm\frac{b}{a}(x - h)$. Substituting $h = 4$, $k = 2$, $a = 6$, $b = 3$, we get $y-2=\pm\frac{3}{6}(x - 4)$, which simplifies to $y-2=\pm\frac{1}{2}(x - 4)$ or $y=\frac{1}{2}x$ and $y=-\frac{1}{2}x + 4$.

Step7: Find the length of the transverse axis

The length of the transverse axis of the hyperbola $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$ is $2a$. Since $a = 6$, the length is $2\times6 = 12$.

Step8: Find the length of the conjugate axis

The length of the conjugate axis of the hyperbola $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$ is $2b$. Since $b = 3$, the length is $2\times3=6$.

Answer:

Center: $(4,2)$
Vertices: $(10,2)$ and $(-2,2)$
Foci: $(4 + 3\sqrt{5},2)$ and $(4-3\sqrt{5},2)$
Eccentricity: $\frac{\sqrt{5}}{2}$
Asymptotes: $y=\frac{1}{2}x$ and $y=-\frac{1}{2}x + 4$
Length of transverse axis: $12$
Length of conjugate axis: $6$