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E3 (adv)

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Saved by PBworks
on June 11, 2007 at 5:05:01 pm
 

E3 (adv) - write the equations of circles and ellipses in transformational form and as mapping rules to visualize and sketch graphs.

 

Lindsay

 

 

Definitions:

 

Circle: a single, closed curved line

 

Ellipse: a single, closed curved line that has one axis longer than the other (x-axis or y-axis) *Example below has a longer y-axis*

 

 

Writing an equation for a circle: To write an equation for a circle, it first needs to be put on a graph. Let's make it easy and have the centre of the circle at the origin of the graph (0,0).

 

The equation of a circle looks like this: [1/r (x-h)]^2 + [1/r (y-k)]^2 = 1  This is called Transformational Form. The r stands for radius, h is the horizontal translation and k is the vertical translation. h and k is also the centre of the circle.

 

The equation for the circle above circle is: [1/5 (x-0)]^2 + [1/5 (y-0)]^2 = 1

 

If we only had the equation and not the diagram of the circle, we would need a mapping rule to make the diagram. A mapping rule looks like this:

(x,y) -> (rx+h, ry+k) Notice how the h and k are now positive? This becomes important later. The r is the radius of the circle, but in a mapping rule the r stands for stretch as well.

 

 

The mapping rule for the above circle is: (x,y) -> (5x+0, 5y+0)

 

 

 

Let's make it harder and change the circle on the graph:

The equation for this circle is: [1/5 (x-0)]^2 + [1/5 (y-3)]^2 =1  All we did with this circle was move it vertically by +3. This +3 is equal to k. This is what happened: [1/5 (x-(0)]^2 + [1/5 (y-(+3)]^2 =1 We did not move the circle horizontally from our last circle so it stays at 0.

 

The mapping rule now is: (x,y) -> (5x+0, 5y+3) In the mapping rule, the h and k are positive, unlike the equation where they are negative. This is to show that the circle moved +3 on the y-axis. It makes it easier to draw.

 

 

 

Writing an equation for an ellipse: The equation for an ellipse is the same as an equation for a circle except we now have to use a stretch. We'll start off with an ellipse with a centre at the origin (0,0).

 

The equation of an ellipse is: [1/a (x-h)]^2 + [1/b (y-k)]^2 =1 This is also called transformational form. In this equation, the only difference from the circle equation is the a and b. Both a and b stand for stretch. The a stands for horizontal stretch (x-axis) and the b stands for vertical stretch (y-axis). The longer of the two stretches is called a major axis, and the smaller stretch is called the minor axis.

 

 

 

The equation for the above ellipse is: [1/3 (x-0)]^2 + [1/6 (y-0)]^2 =1

 

The mapping rule for the above ellipse is: (x,y) -> (3x+0, 6y+0) The radius of x is 3 and the radius of y is 6, so that is what we put for the stretches in the mapping rule.

 

 

Let's make a harder ellipse:

 

So we've done with this ellipse was move it -3 on the x-axis and -3 on the y-axis. The equation for this ellipse would be:

[1/3 (x+3)]^2 + [1/6 (y+3)]^2 =1 All we did was this: [1/3 (x+(-3))]^2 + [1/6 (y+(-3))]^2 =1

 

The mapping rule for this ellipse is: (x,y) -> (3x-3, 6y-3)

 

 

 

On your own: Determine if the following equations are for circles or ellipses, if an ellipses determine the longer and shorter axis (longer-major axis, shorter- minor axis) and then write the mapping rule. Later, try to draw them on a graph.

 

A) [1/5 (x-4)]^2 + [1/5 (y+2)]^2 =1

 

B) [1/120 (x+1)]^2 + [1/2 (y+6)]^2 =1

 

C) (x+5)^2 + [1/8 (y-10)]^2 =1

 

D) [2 (x+4)]^2 + [2 (y-2)]^2 =1

 

 

Answers:

A) Circle, (x,y) -> (5x+4, 5y-2)

B) Ellipse, Major Axis- x: 120, Minor Axis-y: 2, (x,y) -> (120x-1, 2y-6)

C) Ellipse, Major Axis- y: 8, Minor Axis- x: 1, (x,y) -> (x-5, 8y+10)

D) Circle, (x,y) -> (1/2x-4, 1/2y+2)

 

 

 

 

 

 

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