Precalculus Conic Sections

$$\frac{\left(x-3\right)^2}{25}+\frac{\left(y+6\right)^2}{9}=1$$  

$$\frac{\left(x+4\right)^2}{11}+\frac{\left(y+2\right)^2}{64}=1$$

$$\frac{\left(x+2\right)^2}{11}+\frac{\left(y+4\right)^2}{36}=1$$

$$\frac{\left(x-2\right)^2}{25}-\frac{\left(y+4\right)^2}{36}=1$$

$$\frac{\left(x-2\right)^2}{25}+\frac{\left(y+4\right)^2}{9}=1$$  

$$\frac{\left(x-2\right)^2}{16}+\frac{\left(y-4\right)^2}{9}=1$$

$$\frac{\left(x-2\right)^2}{25}-\frac{\left(y+4\right)^2}{9}=1$$

$$\frac{\left(x+2\right)^2}{9}+\frac{\left(y+4\right)^2}{25}=1$$

$$\frac{x^2}{3}-\frac{y^2}{5}=1$$  

$$\frac{x^2}{64}-\frac{y^2}{25}=1$$  

$$\frac{y^2}{4}-\frac{x^2}{16}=1$$  

$$\frac{x^2}{9}-\frac{y^2}{4}=1$$  

They are the cross-sections given when a cone is sliced.

Because it's part of 'da rules'

They are the cross-sections given when any 3D image is sliced.

The minor axis of a vertical ellipse is vertical

The transverse axis of a horizontal hyperbola is horizontal.

Another name for the transverse axis of a hyperbola is the minor axis.

If the equation of a circle is: x² + y² = 8, the diameter of the circle is 16.

The parabola 12y = 6x² -30 opens up

(-1, -4)

(4, -1)

(-1, 6)

(-6, -1)

A parabola has 2 foci.

A circle's diameter is equal to it's radius squared

A hyperbola uses a²+b²=c²

An ellipse uses a²+b²=c²

The distance from the center to a vertex along the transverse axis is 'a' in a hyperbola

$$\frac{\left(y-4\right)^2}{2}-\frac{\left(x-5\right)^2}{7}=1$$   

$$\frac{\left(y-4\right)^2}{4}-\frac{\left(x-8\right)^2}{12}=1$$   

$$\frac{\left(y-5\right)^2}{4}-\frac{\left(x-2\right)^2}{12}=1$$   

$$\frac{\left(y-4\right)^2}{4}-\frac{\left(x-8\right)^2}{12}=1$$   

$$4x^2+2\sqrt[]{3}xy+2y^2-10=0$$  

$$3x^2-14x-y^2-12\sqrt[]{3}y=0$$  

$$5x^2+2\sqrt[]{3}xy+7y^2-16=0$$  

$$4x^2+4\sqrt[]{3}xy+4y^2-4=0$$