Precalculus: Conic Sections Hyperbola Test

$$\frac{\left(x+5\right)^2}{164}-\frac{\left(y-4\right)^2}{41}=1$$

$$\frac{\left(x+5\right)^2}{36}-\frac{\left(y+4\right)^2}{9}=1$$

$$\frac{(y-4)^2}{9}-\frac{(x-5)^2}{36}=1$$

$$\frac{(x+5)^2}{36}-\frac{(y-4)^2}{9}=1$$

1

5

25

16

y = x and y = -x

y = 2x and y = -2x

y = ¹/₂x and y = -¹/₂x

y = 2x and y = 4x

$$\frac{(x-2)^2}{16}-\frac{(y+3)^2}{20}=1$$

$$\frac{(x+3)^2}{16}-\frac{(y-2)}{20}=1$$

$$\frac{(x-2)^2}{20}-\frac{(y+3)^2}{16}=1$$

$$\frac{(y+3)^2}{16}-\frac{(x-2)^2}{20}=1$$

(0, 4)

(4, 0)

(-4, 0)

(0, -4)

$$\frac{\left(x+3\right)^2}{16}-\frac{\left(y+1\right)^2}{4}=1$$

$$\frac{\left(x+3\right)^2}{16}-\frac{\left(y-1\right)^2}{4}=1$$

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

$$\frac{\left(x+3\right)^2}{16}-\left(y+1\right)^2=1$$

$$x^2-y^2+2y-17=0$$

$$x^2+4y^2-8x+8y+16=0$$

$$-3x^2-24x+4y-48=0$$

$$x^2+y^2+4x-14=0$$

foci

center

co-vertices

vertices

(1,-2)

(-1,2)

(-2,1)

(2,-1)