1. **Problem 36:** Given foci at $(-3,-11)$ and $(-3,0)$ and transverse axis endpoints at $(-3,-9)$ and $(-3,-2)$, identify the conic and find its equation. 2. The foci share the same $x$-coordinate, so the conic is vertical. 3. The center is the midpoint of the foci: $$\left(-3, \frac{-11+0}{2}\right) = (-3, -5.5)$$ 4. The distance between the foci is $2c = |0 - (-11)| = 11$, so $$c = \frac{11}{2} = 5.5$$ 5. The transverse axis endpoints are $(-3,-9)$ and $(-3,-2)$, so the length of the transverse axis is $2a = |-2 - (-9)| = 7$, thus $$a = \frac{7}{2} = 3.5$$ 6. For an ellipse or hyperbola, $c^2 = a^2 + b^2$ for hyperbola or $c^2 = a^2 - b^2$ for ellipse. Since $c > a$, this is a hyperbola. 7. Calculate $b^2$: $$b^2 = c^2 - a^2 = 5.5^2 - 3.5^2 = 30.25 - 12.25 = 18$$ 8. The equation of a vertical hyperbola centered at $(h,k)$ is: $$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$ 9. Substitute $h=-3$, $k=-5.5$, $a=3.5$, $b=\sqrt{18}$: $$\frac{(y+5.5)^2}{3.5^2} - \frac{(x+3)^2}{18} = 1$$ --- 1. **Problem 37:** Center at $(-3,6)$, $a=5$, $e=2$, vertical focal axis. 2. Recall eccentricity $e = \frac{c}{a}$, so $$c = ae = 5 \times 2 = 10$$ 3. For a vertical hyperbola, $b^2 = c^2 - a^2 = 10^2 - 5^2 = 100 - 25 = 75$ 4. Equation: $$\frac{(y-6)^2}{25} - \frac{(x+3)^2}{75} = 1$$ --- 1. **Problem 38:** Center at $(1,-4)$, $c=6$, $e=2$, horizontal focal axis. 2. Eccentricity $e = \frac{c}{a}$, so $$a = \frac{c}{e} = \frac{6}{2} = 3$$ 3. For horizontal hyperbola, $b^2 = c^2 - a^2 = 36 - 9 = 27$ 4. Equation: $$\frac{(x-1)^2}{9} - \frac{(y+4)^2}{27} = 1$$