Math 30P Conic(Sections) Conic(General) Conic(Standard) Prerequisite Skills
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The standard form of a circle is:

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The standard form of an ellipse is:
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)
(y  k)^{2} + (x  h)^{2} =
1 (Vertical Ellipse) 
Interactive Activity
A copy of the above applet is provided to study the remainder of the notes for descriptions of stretches.
A copy of the above applet is provided to study the remainder of the notes for descriptions of stretches.

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Standard Form:

Quadratic Standard (Completed Square) Form Interactive Activity Move the sliders for a, h, and k.
A copy of the above applet is provided to study the remainder of the notes for descriptions of stretches.

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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) vertices are at (±a, 0) slope of asymptotes +b/a (y  k)^{2}  (x  h)^{2} =
1 (North  South) vertices are at (0, ±a) slope of asymptotes +a/b 
Interactive Activity
A copy of the above applet is provided to study the remainder of the notes for descriptions of stretches.
A copy of the above applet is provided to study the remainder of the notes for descriptions of stretches.

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 x^{2} + y^{2} = 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 xaxis at (a, 0) and (a, 0). To determine the equation of this ellipse, substitute for x in x^{2} + y^{2} = 1. If you stretch this ellipse vertically by a factor of b, you will obtain an ellipse that crosses the xaxis at (a, 0) and (a, 0) and the yaxis at (0, b) and (0, b). To determine the equation of this ellipse, substitute for y in 
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 xintercepts are ± 8 and the yintercepts are ± 5, or you can find the intercepts algebraically. To find the xintercepts, let y = 0.
The major axis lies along the xaxis; therefore, the ellipse’s vertices (the endpoints of the major axis) are obtained from the xintercepts. The vertices are (8, 0) and (8, 0). The minor axis lies along the yaxis; therefore, the endpoints of the minor axis are obtained from the yintercepts. 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.

Graph x^{2}  y^{2} = 1 on your graphing calculator. What are the asymptotes of the graph? Solution
To display detail near the xintercepts and minimize distortion, use
the following window settings.
The hyperbola x^{2}  y^{2} = 1 crosses the xaxis 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 yaxis 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 x^{2}  y^{2} = 1. Describe and sketch the curve. Solution 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 asymptotes could also be obtained from the original asymptotes,
y = ± x, by applying the same transformations the hyperbola underwent.

Describe the transformations necessary to obtain the graph of from the graph of y^{2}  x^{2} = 1. Describe and sketch the curve. Solution Because is the same as , the graph of y^{2}  x^{2} = 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)

Example 8 Sketch the graph of Solution The centre of the hyperbola is at (3, 2); the transverse axis is parallel to the xaxis; 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.

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 ycoordinate 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. Convert the equation into general form. The general form of the equation is 2y^{2}  x  12y + 20 = 0. 
Example 10 Express x^{2}  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 completedsquare form.

The unit circle, x^{2} + y^{2} = 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. 1. The horizontal stretch factor, a, is 3. The vertical stretch factor, b, is 4.
2. The horizontal translation, k, is 3.
The general form of the equation is 16x^{2} + 9y^{2}  96x = 0. 

Parts of this work has been adapted from a Math 30 Pure learning resource originally produced and owned by Alberta Education
Math 30P Conic(Sections) Conic(General) Conic(Standard) Prerequisite Skills