### 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 **2** **stretches**. 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 all 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 ellipse 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)

## Comments (1)

## Anonymous said

at 7:22 pm on Jun 14, 2007

Great Job Lindsay!

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