]> Chapter 10 Notes

10: Quadratic Relations

Perhaps this chapter would be better titled Conic Sections…oh well.

I give these notes in the order that the text presents them—however, we will probably skip around a bit in class. Also, I'll be presenting these ideas as I understand them—and I hope that this will make it easier than the standard (textbook) approach.

10-1: Conic Sections

A conic section is a plane curve that can be constructed by intersecting a plane and double-napped cone (looks like two cones put tip-to-tip). When the plane is perpendicular to the axis of the cone, the curve is a circle. As the angle between the plane and the cone's axis is decreased, other shapes are created (ellipse and parabola) until the plane and conic axis are parallel (hyperbola). Here's a diagram:

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Naturally, you know what circles and parabolas looks like. Here are an ellipse and hyperbola:

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Examples

There aren't any problems worth doing in this section!

10-2: Parabolas

Parabolas are in familiar territory, as we looked at them back in Chapter 5. This time, we will consider parabolas in both directions: up/down and left/right.

The up/down parabola that you know and love has the equation $y = x^{2}$ . The left/right parabola that you will come to know and love has the equation $x = y^{2}$ . Unfortunately, not all parabolas are this simple! You already know how to move the vertical parabola around: $y = a {(x - c)}^{2} + d$ . The horizontal parabola looks like $x = \frac{1}{b} {(y - d)}^{2} + c$ . To use this, divide the x-values by b; shift d units up and c units right.

The text defines a parabola as the locus (set) of points that are equidistant from a point (the focus) and a line (the directrix). While interesting, I do not plan to spend any time attacking parabolas with this approach.

Examples

[1.] Sketch $x = 2 {(y - 1)}^{2} + 3$ .

Sketch a standard sideways parabola; multiply the x-values by 2; shift one up and three right.

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[2.] Sketch $y = x^{2} + 6 x + 7$ .

I'll need to complete the square on this one!

$x^{2} + 6 x + 7 \Rightarrow (x^{2} + 6 x + 9) - 2 \Rightarrow {(x + 3)}^{2} - 2$

Take a standard parabola; shift down three and left two.

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10-3: Circles

Circles have the equation $x^{2} + y^{2} = r^{2}$ , where r is the radius of the circle. Also, note the alternate definition: the locus of points equidistant from a point (the center).

Not all circles are centered at the origin! The equation ${(x - c)}^{2} + {(y - d)}^{2} = r^{2}$ is a circle of radius r centered at $(c, d)$ .

Examples

[3.] Write the equation of a circle with radius 3 and center $(- 1, 4)$ .

${(x + 1)}^{2} + {(y - 4)}^{2} = 9$

[4.] A circle of radius 2, centered at the origin, is moved three units left and up five units. Write the equation of the new circle.

${(x + 3)}^{2} + {(y - 5)}^{2} = 4$

[5.] Find the equation of the circle shown below.

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The center is at $(- 2, 3)$ and the radius is 4, so the equation is ${(x + 2)}^{2} + {(y - 3)}^{2} = 16$ .

[6.] Find the center and radius of the circle with equation $x^{2} + 2 x + y^{2} - 4 y + 5 = 10$ .

This will require completing the square.

$(x^{2} + 2 x + 1) + (y^{2} - 4 y + 4) = 10 - 5 + 1 + 4 \Rightarrow {(x + 1)}^{2} + {(y - 2)}^{2} = 10$

10-4: Ellipse

An ellipse has the equation $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ . The bigger denominator tells you the direction of the ellipse—if the denominator under x is bigger, then the ellipse is larger horizontally. I think that this is simpler than memorizing two equations!

The alternate definition of an ellipse: the locus of points where the sum of distances to two fixed points (the foci) is constant.

To graph the ellipse, plot the points $(a, 0)$ , $(- a, 0)$ , $(0, b)$ and $(0, - b)$ . Connect these points in a nice smooth curved line.

The longer of the segments $(a, 0)$ , $(- a, 0)$ and $(0, b)$ , $(0, - b)$ is called the major axis. The smaller segment is called the minor axis. The endpoints of these segments are the vertices of the ellipse.

The axes of an ellipse are like a diameter of a circle; a semi-axis of an ellipse is like a radius of a circle—and must have as its value either a or b!

Examples

[7.] Find the equation of the ellipse shown below.

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The center is at $(2, 1)$ ; the major semi-axis (vertical) is 5, and the minor semi-axis is 4. The equation of the ellipse is $\frac{{(x - 2)}^{2}}{16} + \frac{{(y - 1)}^{2}}{25} = 1$ .

[8.] Sketch $\frac{x^{2}}{4} + \frac{y^{2}}{25} = 1$ .

This is centered at the origin; the major semi-axis (vertical) is 5, and the minor semi-axis is 2. That puts the vertices at $(0, 5)$ , $(0, - 5)$ , $(2, 0)$ and $(- 2, 0)$ .

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[9.] The vertices of an ellipse are $(1, 0)$ , $(- 1, 0)$ , $(0, \sqrt{3})$ and $(0, - \sqrt{3})$ . Write the equation of the ellipse.

These points are centered at the origin. This is a vertical ellipse, since the vertical axis is longer. The major semi-axis is $\sqrt{3}$ , and the minor semi-axis is 1.

The equation is $\frac{x^{2}}{1} + \frac{y^{2}}{3} = 1$ .

10-5: Hyperbolas

There are two varieties of hyperbola—vertical and horizontal. The horizontal hyperbola has equation $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , and the vertical hyperbola is $\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ .

The alternate definition of a hyperbola: the locus of points where the difference in the distances to two points (the foci) is constant.

To graph the hyperbola, plot the points $(a, b)$ , $(- a, b)$ , $(a, - b)$ and $(- a, - b)$ . Connect these points with a light, dotted rectangle. Connect the corners with dashed lines—these will be the asymptotes. Draw the two branches of the hyperbola in the direction that the equation indicates, approaching the asymptotes and touching the dotted rectangle.

The points where the hyperbola touches the dotted rectangle are called the vertices. The segment connecting the vertices is called the transverse axis.

Examples

[10.] Find the equation of the hyperbola shown below.

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This is a horizontal hyperbola. Its vertices are at $(- 3, 0)$ and $(3, 0)$ , which tells where the left and right edges of the rectangle are. Note that from the vertices, the asymptotes are up (or down) two units. This tells us the values of a and b— $a = 3$ and $b = 2$ . The equation of the hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ .

[11.] Sketch the graph of $x^{2} - 9 y^{2} = 9$ .

Just divide to get this into standard form: $\frac{x^{2}}{9} - \frac{y^{2}}{1} = 1$ . This is horizontal, centered at the origin. Plot $(3, 1)$ , $(- 3, 1)$ , $(3, - 1)$ and $(- 3, - 1)$ ; connect the corners to make the asymptotes. The vertices are at $(3, 0)$ and $(- 3, 0)$ .

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10-6: Translating Conic Sections

So far, I've mostly written about the standard conics—centered at the origin. Naturally, this isn't the only way.

You already know how to move the parabolas around…

The general ellipse looks like $\frac{{(x - c)}^{2}}{a^{2}} + \frac{{(y - d)}^{2}}{b^{2}} = 1$ . This shifts the center to coordinates $(c, d)$ —graph the ellipse as described above, then just shift the entire thing over to the new center. Note that a circle is just a special ellipse, where the denominators are equal.

The hyperbolas look like $\frac{{(x - c)}^{2}}{a^{2}} - \frac{{(y - d)}^{2}}{b^{2}} = 1$ and $\frac{{(y - d)}^{2}}{b^{2}} - \frac{{(x - c)}^{2}}{a^{2}} = 1$ . Again, this simply shifts the center—graph the hyperbola as described earlier, then shift everything over.

The really fun thing about this is completing the square in order to find the new center!

Examples

[12.] Sketch the graph of $y = - 3 x^{2} - 12 x - 14$ .

Since only one variable is squared, this must be a parabola—and since x is squared, it must be a vertical parabola.

$- 3 x^{2} - 12 x - 14 = - 3 (x^{2} + 4 x) - 14 = - 3 (x^{2} + 4 x + 4) - 14 + 12 = - 3 {(x + 2)}^{2} - 2$

Flip a standard parabola over the x-axis and multiply the y-values by 3; shift two left and two down.

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[13.] Sketch the graph of $9 x^{2} - 36 x - y^{2} + 2 y + 35 = 9$ .

Since the signs on the squared terms are different (one positive and one negative), this must be a hyperbola. I'll need to complete the square for both variables.

$9 (x^{2} - 4 x) - (y^{2} - 2 y) + 35 = 9 \Rightarrow 9 (x^{2} - 4 x + 4) - (y^{2} - 2 y + 1) + 35 = 9 + 36 - 1$

$\Rightarrow 9 {(x - 2)}^{2} - {(y - 1)}^{2} = 9 \Rightarrow \frac{{(x - 2)}^{2}}{1} - \frac{{(y - 1)}^{2}}{9} = 1$

This is a horizontal hyperbola. Before shifting the center, the rectangle corners are at $(1, 3)$ , $(1, - 3)$ , $(- 1, 3)$ and $(- 1, - 3)$ , and the vertices are at $(1, 0)$ and $(- 1, 0)$ . Shifting the center to $(2, 1)$ puts the rectangle corners at $(3, 4)$ , $(3, - 2)$ , $(1, 4)$ and $(1, - 2)$ , with vertices at $(3, 1)$ and $(1, 1)$ .

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[14.] Sketch the graph of $9 x^{2} - 18 x + 4 y^{2} + 16 y + 25 = 36$ .

Since both variables are squared and positive, this is an ellipse.

$9 (x^{2} - 2 x) + 4 (y^{2} + 4 y) = 11 \Rightarrow 9 (x^{2} - 2 x + 1) + 4 (y^{2} + 4 y + 4) = 11 + 9 + 16$

$\Rightarrow 9 {(x - 1)}^{2} + 4 {(y + 2)}^{2} = 36 \Rightarrow \frac{{(x - 1)}^{2}}{4} + \frac{{(y + 2)}^{2}}{9} = 1$

This is a vertical ellipse. Before shifting, the vertices are at $(0, 3)$ , $(0, - 3)$ , $(2, 0)$ and $(- 2, 0)$ . After shifting the center to $(1, - 2)$ , the vertices are at $(1, 1)$ , $(1, - 5)$ , $(3, - 2)$ and $(- 1, - 2)$ .

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Page last validated 2010-08-15