Conics.

Hello, Anemori!

Quote:

a. Produce a conic system of inequalities representing the "doughnut-shape"
created by an ellipse inside a circle.

Code:

| ..* * *.. *:::::|:::::* *:::::::|:::::::* *::::::::|::::::::* :::::::::**:::::::: *::::* |b *::::* c - - *:-:* - - + - - *:-:* - - - *::::* | a *::::* :::::::***:::::::: *::::::::|::::::::* *:::::::|:::::::* *:::::|:::::* * * * |

The equation of the ellipse is: . $\frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$

The equation of the circle is: . $x^2 + y^2 \:=\:c^2,\,\text{ where }c \,\geq\,a,b$

System: . $\begin{Bmatrix} \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} & \geq & 1 \\ \\[-3mm]x^2 + y^2 & \leq & c^2 \end{Bmatrix}$

Quote:

b. Produce a conic system that has exactly two points of intersection
and consists of a circle and ellipse with the same area.

The equation of an ellipse is: . $\frac{x^2}{a^2} + \frac{y^2}{b^2} \:=\:1$
. . Its area is: . $\pi ab$

The equation of a circle is: . $x^2 + y^2 \:=\:r^2$
. . Its area is: . $\pi r^2$

The areas are equal: . $\pi ab \:=\:\pi r^2 \quad\Rightarrow\quad ab \:=\:r^2$

Let: . $a = 4,\;b = 1,\;r = 2$

The ellipse is: . $\frac{x^2}{16} + \frac{y^2}{1} \;=\;1$

Code:

1| o o o o | o o | o - o - - - - - + - - - - - o - - o | o 4 o | o o o o |

The circle is: . $x^2 + y^2 \:=\:4$

Code:

2| ◊ ◊ ◊ ◊ | ◊ ◊ | ◊ | - - ◊ - - - + - - - ◊ - - | 2 ◊ | ◊ ◊ | ◊ ◊ ◊ ◊ |

Move the circle 2 units to the right.

Code:

| : | ◊ ◊ ◊ o o ♥ : ◊ o |◊ o ◊ o | : o - - o - - - - - ◊ - - - + - o - ◊ - - o | : o o |◊ o ◊ o o ♥ : | ◊ ◊ ◊

System: . $\begin{Bmatrix}\dfrac{x^2}{16} + \dfrac{y^2}{1} \:=\:1 \\ \\[-3mm] (x-2)^2 + y^2 \:=\:4 \end{Bmatrix}$