Mathematics, 28.06.2019 02:30, LarryJoeseph

Use the graph below to write 5 conic section equations that create it. use desmos to check if your equations produce this picture. hint: 3 circles, 1 ellipse and 1 parabola 1.the head(red circle) 2. the left eye (blue circle) 3. the right eye (green circle) 4. the mouth (purple ellipse) 5. the body (black parabola)

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answered: rainbowsauxe

Parabola is a plane curve. The iconic character can be made with the following equation as given below,

$(x-0)^{2}+(y-0)^{2}=6^{2}$ $(x+2)^{2}+(y-2)^{2}=1^{2}$ $(x-2)^{2}+(y-2)^{2}=1^{2}$ $(\dfrac{x}{3})^{2}+(y+2)^{2}-1=0$ $y=-(x-0)^{2}-6$

We need to draw 3 circles, an ellipse and a parabola, in order to get the desired digram.

For a circle, we only need to know the radius of the circle and the center of the circle,

1. The head(red circle)

The head of the circle is drawn from the origin of the coordinate, while the radius can be found by dividing the length between the points where the circle cuts the x-axis.

$Radius =\dfrac{6-(-6)}{2} = 6$

Equation of the circle face,

$\text{Equation of the circle face} = {(x-0)^2+(y-0)^2} = 6^2$

2. The Left Eye (Blue Circle)

3. The Right Eye (Green Circle)

Similarly for the eyes of the diagram,

The right eye = $(x-2)^{2}+(y-2)^{2}=1^{2}$

The left eye = $(x+2)^{2}+(y-2)^{2}=1^{2}$

4. The Mouth (Purple Ellipse)

The equation of the parabola can be written as,

(x/3)² +(y+2)² -1 = 0

5. The Body (Black Parabola)

The parabola can be made using the equation,

Equation of a parabola

y = a(x-h)² + k

where,

(h, k) are the coordinates of the vertex of the parabola in form (x, y);

a defines how narrower is the parabola, and the "-" or "+" that the parabola will open up or down.

Therefore, the equation for the body,

Equation of the parabola = y = -(x-0)^2 -6

Learn more about Parabola:

link

Use the graph below to write 5 conic section equations that create it. use desmos to check if your e

answered: barisegebalci165

See below for the graph.

Step-by-step explanation:

A circle or ellipse can be defined using the same sort of equation. Here, we have chosen to use the formulation ...

... ((x -a)/p)^2 +((y -b)/q)^2 -1 = 0

This will be the general form of the equation for an ellipse with center (a, b) and semi-axes p and q, in the x- and y-directions, respectively. When the axes are the same length, the ellipse is a circle.

By defining the function ...

... c(a, b, p, q, x, y) = ((x -a)/p)^2 +((y -b)/q)^2 -1

we can use the same function for all of the circles/ellipses in the figure. The parabola has vertex (0, -6) and a vertical scale factor of -1, so it can be formulated using the vertex form:

... y = k(x -a)^2 +b . . . . . for vertex (a, b) and vertical scale factor k.

_____

The equations

(x/6)² +(y/6) -1 = 0(x+2)² +(y-2)² -1 = 0(x-2)² +(y-2)² -1 = 0(x/3)² +(y+2)² -1 = 0y = -x² -6
Use the graph below to write 5 conic section equations that create it. use desmos to check if your e