Standard Form

The standard form of a circle is:

(x - h)² + (y - k)² = r²

where (h, k) is the centre of the circle.

Interactive Activity

• Click on the applet to activate
• .Move the h, k and r sliders. Observe the change(s) have on the position and size of the circle.

The standard form of an ellipse is:

(x - h)² + (y - k)² = 1, where:
   a²            b²

In this resource, a represents the horizontal stretch factor and b represents the vertical stretch factor.

Note: In some resources (Alberta Distance Learning Math 30 Pure correspondence course), a represents the stretch factor in the direction of the major(longest) axis and b represents the stretch factor in the direction of the minor axis.

(x - h)² + (y - k)² = 1 (Horizontal Ellipse)
  a²            b²

(y - k)² + (x - h)² = 1 (Vertical Ellipse)
   a²            b²

Interactive Activity

• Click on the applet to activate
• .Alternate between the a, b, h and k sliders. Observe the change(s) have on the position and shape of the ellipse..

Standard Form:

	East - West x = a(y - k)2 + h	North - South y = a(x - h)2 + k
vertex	(h, k)	(h, k)
transformed function??	Vertical stretch by factor of a x = af(y) Vertical translation of k x = f(y - k) Horizontal translation of h x = f(y) +h	Horizontal stretch by a factor of a y = af(x) Vertical translation of k y = f(x) + k Horizontal translation of h y = f(x - h)
horizontal stretch factor	a	None
vertical stretch factor	None	a
domain - set of possible x-values	If a > 0, opens right y > k If a < 0, opens left y < k	x R
range - set of possible y-values	y R	If a > 0, opens up x > h If a < 0, opens down x < h
axis of symmetry	y = k	x = h
Algebraic Description of Transformation	horizontal stretch: x x/a: Replace the x in the original function with x/a. If a<0, the graph will also be a reflection about the y-axis. horizontal translation: x x - h: Replace the x in the original function with (x-h) vertical translation: y y - h: Replace the y in the original function with (y-k)	vertical stretch: y y/a: Replace the y in the original function with y/a. If a<0, the graph will also be a reflection about the x-axis horizontal translation: x x - h: Replace the x in the original function with (x-h) vertical translation: y y - h: Replace the y in the original function with (y-k)
Mapping notation to find location of data points after a transformation.	stretch: (x, y) (x/a, y) translation: (x, y) (x+h, y+k)	stretch: (x, y) (x, y/a) translation: (x, y) (x+h, y+k)
	East - West	North - South

Quadratic Standard (Completed Square) Form Interactive Activity

Move the sliders for a, h, and k.

• Study the effect of changing a from positive to negative values.
• Study the effect changing h and k has on the vertex (the minimum/maximum value) position.
• Find the x-intercept(s) on the graph.
• Find the y-intercept(s) on the graph.

Linked Source - Ron Blond

A copy of the above applet is provided to study the remainder of the notes for descriptions of stretches.

Linked Source - Ron Blond

Standard Form

In this resource, a represents the horizontal stretch factor and b represents the vertical stretch factor.

Note: In some resources a represents the stretch factor in the direction of the transverse (major) axis and b represents the stretch factor in the direction of the conjugate (minor) axis.

(x - h)² - (y - k)² = 1 (East - West)
    a²           b²

vertices are at (±a, 0)

slope of asymptotes +b/a

(y - k)² - (x - h)² = 1 (North - South)
   a²            b²

vertices are at (0, ±a)

slope of asymptotes +a/b

Interactive Activity

• Click on the applet to activate
• .Move the a, b, h and k sliders. Observe the change(s) have on the position and size of the hyperbola.
• To form the asymptotes easily on the graph, form a rectangle using a and b.

Example 1

The equation of the circle in standard form is:

(x - 2)² + (y - 3)² = 26.

Write the equation in general form.

Solution

Example 2

Think of an ellipse that is formed by stretching a circle horizontally and/or vertically. For example, the circle x² + y² = 1 is centred at the origin with a radius of 1 unit. If you stretch it horizontally by a factor of a, you will obtain an ellipse that crosses the x-axis at (-a, 0) and (a, 0). To determine the equation of this ellipse, substitute for x in x² + y² = 1.

If you stretch this ellipse vertically by a factor of b, you will obtain an ellipse that crosses the x-axis at (-a, 0) and (a, 0) and the y-axis at (0, b) and (0, -b). To determine the equation of this ellipse, substitute for y in

Example 3

Use your graphing calculator to graph the following ellipse; then determine the intercepts, the vertices, the endpoints of the minor axis, and the domain and range.

Solution

Use a square display to obtain a graph with mimimal distortion:

[ZOOM], 5:Zsquare

First, solve the equation for y so you can enter this equation in the form y = f(x).

You can use the Table feature on your calculator to confirm that the x-intercepts are ± 8 and the y-intercepts are ± 5, or you can find the intercepts algebraically.

To find the x-intercepts, let y = 0.

To find the y-intercepts, let x = 0.

The major axis lies along the x-axis; therefore, the ellipse’s vertices (the endpoints of the major axis) are obtained from the x-intercepts. The vertices are (8, 0) and (-8, 0).

The minor axis lies along the y-axis; therefore, the endpoints of the minor axis are obtained from the y-intercepts. The endpoints of the minor axis are (0, 5) and (0, -5).

Using the intercepts, the domain is -8 x 8 and the range is -5 y 5.

Example 4

Express the following equation in general form.

Draw the graphs of the unit circle and the ellipse.What is the domain and range of the ellipse?

Solution

Graphs of the unit circle and the ellipse.

The horizontal axis extends from (0, 4) to (6, 4). Therefore, the domain of the ellipse is 0 x 6.

The vertical axis extends from (3, 2) to (3, 6). Therefore, the range of the ellipse is 2 y 6.

Example 5

Graph x² - y² = 1 on your graphing calculator. What are the asymptotes of the graph?

Solution
Express x² - y² = 1 in the form y = f(x).

To display detail near the x-intercepts and minimize distortion, use the following window settings.

As you go outward from the origin, the branches of the hyperbola appear to approach these lines. Consider y = x and the portion of the hyperbola in the first quadrant obtained from

The vertical distance between the two graphs is

Graph this difference on your graphing calculator using the following window settings.

Check the table of values to verify that this difference approaches 0 as x increases.

Using this method, you can verify that both branches of the hyperbola lie between and approach lines y = x and y = -x.

The hyperbola x² - y² = 1 crosses the x-axis at (-1, 0) and (1, 0). These points on the horizontal axis of symmetry are the vertices of this hyperbola. The axis of symmetry that joins the vertices is called the transverse axis. The vertical axis of symmetry, which lies along the y-axis in this case, is perpendicular to the transverse axis and is called the conjugate axis.

The two axes of the hyperbola intersect at the centre of the hyperbola. In this case the centre is at the origin.

Example 6

Describe the transformations necessary to obtain the graph of the following hyperbola from the graph of x² - y² = 1. Describe and sketch the curve.

Solution
Because is the same as , the graph of x² - y² = 1 is stretched horizontally by a factor of 4 and vertically by a factor of 3.

Because the hyperbola opens to the left and to the right, use the horizontal stretch factor of 4 to locate the vertices. (-1 × 4, 0) = (-4, 0) and (1 × 4, 0) = (4, 0)
The slopes of the asymptotes of x² - y² = 1, using . Therefore, the slopes of the new asymptotes are

Because the asymptotes pass through the origin, their equations are of the form y = mx. Thus, the asymptotes are .

The asymptotes could also be obtained from the original asymptotes, y = ± x, by applying the same transformations the hyperbola underwent.

Sketch the graph by first locating the vertices. Then draw the asymptotes using their slopes. Finally, sketch the branches of the hyperbola opening to the left and to the right from the vertices and between the asymptotes.

Example 7

Describe the transformations necessary to obtain the graph of from the graph of y² - x² = 1. Describe and sketch the curve.

Solution

Because is the same as , the graph of y² - x² = 1 is stretched horizontally by a factor of 3 and vertically by a factor of 2.

Because the hyperbola opens upward and downward, use the vertical stretch factor of 2 to locate the vertices.

(0, -1 × 2) = (0, -2) and (0, 1 × 2) = (0, 2)
The slopes of the asymptotes of y² - x² = 1, using are Therefore, the slopes of the new asymptotes are

Because the asymptotes pass through the origin, their equations are of the form y = mx. Thus, the asymptotes are

Sketch the graph by first locating the vertices. Then draw the asymptotes using their slopes. Finally, sketch the branches of the hyperbola opening upward and downward from the vertices and between the asymptotes.

Example 8

Sketch the graph of

Solution

The centre of the hyperbola is at (3, 2); the transverse axis is parallel to the x-axis; and the hyperbola opens to the left and to the right.

Because a = 5, the vertices are 5 units to the left and right of the centre, (3, 2). Therefore, the vertices are at (3 - 5, 2) = (-2, 2) and (3 + 5, 2) = (8, 2).

The slopes of the asymptotes are

Sketch the graph by first locating the centre and the vertices. Then draw the asymptotes from their slopes. Finally, sketch the branches of the hyperbola opening to the left and to the right from the vertices and between the asymptotes.

Example 9

In standard form and general form, determine the equation of a parabola and the line of symmetry with its vertex at (2, 3) with a horizontal axis of symmetry, and passing through (10, 5). Graph the parabola.

Solution

Because the axis of symmetry is horizontal, the standard form of the equation of this parabola is x - h = a(y - k)². The axis of symmetry passes through the vertex so the y-coordinate of the axis of symmetry.

The vertex is (2, 3); so, h = 2, k = 3 and the axis of symmetry is y = 3.

x - h = a(y - k)²

x - 2 = a(y - 3)²

Next, find a. Because the graph passes through (10, 5), substitute x = 10 and y = 5 into the equation.

The standard form of the equation is x - 2 = 2(y - 3)².

Convert the equation into general form.

The general form of the equation is 2y² - x - 12y + 20 = 0.

Example 10

Express x² - 4x + 2y - 8 = 0 in standard form. Then state the vertex, axis of symmetry, and direction of opening; and sketch the graph.

Solution

Note: As for the other conics, the standard form is also called completed-square form.

Remember: Always end up with one of these forms when completing the square:
y - k = a(x - h)² or x - h = a(y - k)²

Example 11

The unit circle, x² + y² = 1, is stretched horizontally by a factor of 3, vertically by a factor of 4, and then translated 3 units to the right. The result is an ellipse. The equation in general form of this ellipse is ____.

Solution

(x - h)² + (y - k)² = 1 is the standard from for an ellipse.
    a²           b²

1. The horizontal stretch factor, a, is 3. The vertical stretch factor, b, is 4.

(x)² + (y)² = 1 is the standard from for an ellipse.
 a²       b²

(x)² + (y)² = 1 is the standard from for an ellipse. Multiply both sides by 144.
 3²      4²

16x² + 9y² = 144

2. The horizontal translation, k, is 3.

16(x - 3)² + 9y² = 144

Convert the equation into general form.

16(x - 3)² + 9y² = 144

16x² - 96x + 144 + 9y² = 144

16x² + 9y² - 96x = 0

The general form of the equation is 16x² + 9y² - 96x = 0.