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You CAN Ace Calculus

17calculus > conics

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Conics

A conic (or conic section) is a smooth curve formed when a plane intersects a pair of right circular cones placed point-to-point. The angle of the plane measured with respect to the axis running through the point of the cones, determines the type of conic that is formed.

### Search 17Calculus

There are three types of curves.
1. Parabolas
2. Ellipses (circles are special cases of ellipses and are sometimes listed as a fourth type)
3. Hyperbolas

Parabolas and hyperbolas are very similar and are easy to confuse. One difference is that there are a pair of curves in the case of a hyperbola but parabolas occur as a single curve.

The general equation for all these equations is $$Ax^2+Bxy+Cy^2+$$ $$Dx+Ey+F=0$$. There are a lot of differences in this equation for each curve. Let's discuss each type.

 Parabola

A parabola is formed when a plane intersections a cone parallel to the side of the cone. This intersection produces only one curve (as compared to two in a hyperbola).

Depending on the orientation of the cone with respect to the coordinate axis (xy-axis in 2 dimensions), the equations will be different. We will look only at the two cases where the coordinate axes runs parallel to the axis of the cone and perpendicular to the axis of the cone. These two cases will produce four possible parabolas.
1. Parabola opens up.
2. Parabola opens down.
3. Parabola opens to the left.
4. Parabola opens to the right.

We can combine types 1 and 2 to get one equation form and similarly, types 3 and 4 can be combined for a second equation form.

$$(x-h)^2 = 4p(y-k)$$ opens up or down opens left or right

Of course, as with many mathematical equations, there are several ways to write this. However, written this way, we can directly pull a lot of information from it for the graph. (It is easy to convert from the form $$y=ax^2+bx+c$$ for example that you are probably more familiar with, using some algebra and completing the square. )

We use special terms to describe parts of the parabola. You are probably familiar with the vertex. The other attributes are listed below (with equations from a parabola that opens up or down; you should be able to translate them to the other orientation) and shown in the figure on the right. Shown very well in this figure, are two pink lines with a black bar crossing them. The black bar indicates that the length of these lines are equal to one another. That is the definition that makes the parabola a parabola.
Classify - - When looking at the equation in the form $$Ax^2+Bxy+Cy^2+$$ $$Dx+Ey+F=0$$, a parabola will have $$B=0$$ and either $$A=0$$ or $$C=0$$ but not both.

$$(x-h)^2 = 4p(y-k)$$

vertex

$$(h,k)$$

axis of symmetry

$$x=h$$

focus

$$(h,k+p)$$

directrix

$$y=k-p$$

$$Ax^2+Dx+Ey+F=0$$

Example $$(x-2)^2=4(y-3)$$

$$(y-k)^2 = 4p(x-h)$$

vertex

$$(h,k)$$

axis of symmetry

$$y=k$$

focus

$$(h+p,k)$$

directrix

$$x=h-p$$

$$Cy^2+Dx+Ey+F=0$$

Example $$(x-2)=-4(y-3)^2$$

Practice 1

Sketch the graph of the parabola $$\displaystyle{ \frac{x^2}{12} = \frac{y}{3} }$$.

solution

Practice 2

Find the focus and the directrix of the parabola $$\displaystyle{ y = \frac{2}{5}x^2 }$$.

solution

Practice 3

Write the standard form of the equation for the parabola with the vertex at the origin and the focus at $$(1/8,0)$$.

solution

Practice 4

Write the standard form of the equation for the parabola with the vertex at the origin and the directrix at $$y=-5/6$$.

solution

Practice 5

Classify and list the attributes of the conic $$x^2-4x-4y=0$$.

solution

 Ellipse

An ellipse is formed when a plane crosses that is not parallel to one of sides of the cone and not parallel to the axis of the cone. A circle is a special form of an ellipse where the plane is perpendicular to the axis of the cone.

The standard equation for an ellipse is $$\displaystyle{ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 }$$. The plot on the right contains more information than we need right now but it will suffice. The longer axis is called the major axis (in this plot it is horizontal). The shorter axis is called the minor axis. The vertices are located on the ellipse where it crosses the major axis. The foci are also on the major axis, labeled F1 and F2 on this plot.

The major axis is determined by the denominators, $$a^2$$ and $$b^2$$. The larger value is in the denominator of the major axis, i.e. if $$a > b$$ then the major axis is parallel to the x-axis. We need to define a value c where $$c^2=\abs{a^2-b^2}$$ which will help us determine the location of the foci.

These tables contain the main attributes of an ellipse. We assume here that $$a > b$$. Similar equations exist for $$a < b$$.

$$\displaystyle{ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 }$$

center

$$(h,k)$$

major axis

$$y=k$$

vertices

$$(h \pm a, k),$$ $$(h, k \pm b)$$

foci

$$(h \pm c,k)$$

$$c^2=\abs{a^2-b^2}$$

eccentricity

$$e=c/a$$

$$\displaystyle{ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 }$$

center

$$(h,k)$$

major axis

$$x=h$$

vertices

$$(h, k \pm a),$$ $$(h \pm b, k)$$

foci

$$(h,k \pm c)$$

$$c^2=\abs{a^2-b^2}$$

eccentricity

$$e=c/a$$

Notes
1. The eccentricity e is not the same as the irrational constant e. Since the foci are closer to the center than the vertices, it follows that $$c < a$$ and therefore $$0 < e < 1$$.
2. Notice in the standard form of the equation, both terms are positive. This is how you know the graph is an ellipse and not a hyperbola.
3. In the general form of the equation, $$Ax^2+Bxy+Cy^2+$$ $$Dx+Ey+F=0$$, $$A > 0$$ and $$C > 0$$.

Okay, time for some fun videos about ellipses. Here are a couple of videos about playing pool on an elliptical table. They clearly show the relationship between the foci and demonstrate some fun physics at the same time.

 Numberphile - Elliptical Pool Table

Practice 6

Write the standard form of the equation for an ellipse, centered at the origin, vertical major axis of length 8 and minor axis of length 2.

solution

Practice 7

Write the standard form of the equation for an ellipse, centered at the origin, with x-intercepts at $$\pm 12$$ and foci at $$(0,\pm 5)$$.

solution

Practice 8

Write the standard form of the equation for an ellipse, centered at the origin, with minor axis of length 6 and foci at $$\pm 8, 0$$.

solution

Practice 9

Find the intercepts of the ellipse $$\displaystyle{\frac{y^2}{100}+\frac{x^2}{121}=1}$$.

solution

Practice 10

Graph the ellipse $$\displaystyle{1-\frac{y^2}{16}=x^2}$$.

solution

Practice 11

Find the center and the radius of the circle $$x^2+2x+y^2=4$$.

solution

Practice 12

Sketch the circle $$x^2+y^2=4x$$.

solution

Practice 13

Sketch the circle $$x^2+y^2+6y=0$$.

solution

Practice 14

Sketch the circle $$x^2+y^2+2x+2y=2$$.

solution

Practice 15

Sketch the circle $$x^2+y^2+10x-20y+100=0$$.

solution

Practice 16

Sketch the circle $$2x^2+2y^2+2x-2y=1$$.

solution

Practice 17

Sketch the circle $$9x^2+9y^2-6x-12y=11$$.

solution

Practice 18

Graph $$\displaystyle{\frac{x^2}{9}+\frac{y^2}{5}=1}$$.

solution

Practice 19

Graph $$4x^2+y^2=16$$.

solution

Practice 20

Graph $$x^2+2y^2-6x+4y+7=0$$.

solution

 Hyperbola

The hyperbola is the most complicated of the three and, consequently, the most interesting. Again, the plot on the right contains more information than we need right now, but it will suffice.
The vertices ($$\pm a$$ on this plot) and the foci (F1 and F2 on this plot) lie on the transverse axis with the center at $$(h,k)$$. The standard equations are

 $$\displaystyle{ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 }$$ horizontal transverse axis $$\displaystyle{ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 }$$ vertical transverse axis

As we did with the ellipse, we define an intermediate value $$c^2=a^2+b^2$$ which will help us locate the foci.
For a hyperbola, we need to know the equations of the lines in blue on the plot. These are asymptotes.
For a horizontal transverse axis, the asymptotes are $$\displaystyle{ y=k \pm \frac{b}{a}(x-h) }$$.
For a vertical transverse axis, the asymptotes are $$\displaystyle{ y=k \pm \frac{a}{b}(x-h) }$$.
Similar to the ellipse, we define the eccentricity as $$e=c/a$$. The results are summarized next.

$$\displaystyle{ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 }$$

horizontal transverse axis

center

$$(h,k)$$

vertices

$$(h \pm a, k)$$

foci

$$(h \pm c,k)$$

asymptotes

$$\displaystyle{ y=k \pm \frac{b}{a}(x-h) }$$

$$c^2=a^2+b^2$$

eccentricity

$$e=c/a$$

$$\displaystyle{ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 }$$

vertical transverse axis

center

$$(h,k)$$

vertices

$$(h, k \pm a)$$

foci

$$(h,k \pm c)$$

asymptotes

$$\displaystyle{ y=k \pm \frac{a}{b}(x-h) }$$

$$c^2=a^2+b^2$$

eccentricity

$$e=c/a$$

Notes
1. Notice that the value of c is different here than for an ellipse.
2. Since $$c > a$$, $$e > 1$$.

The hyperbola is quite a complicated graph with lots of features. Here is a video that goes into more detail.

 MIP4U - Conic Sections: The Hyperbola part 1 of 2

Okay, time for some practice problems on hyperbolas. Although it may seem like we have given you a lot of information on this page, we have just skimmed the surface in the discussion of these three figures, especially the hyperbola. There are a lot more interesting features that we hope you get to explore in your class.

Practice 21

Sketch the graph of the hyperbola $$\displaystyle{ (y+4)^2 - \frac{x^2}{25} = 1 }$$.

solution

Practice 22

Graph $$\displaystyle{\frac{x^2}{49}-\frac{y^2}{25}=1}$$.

solution

Practice 23

Write the equation of the hyperbola that has vertices $$(-2,-5), (4,-5)$$ and foci $$(-4,-5), (6,-5)$$.

solution

Practice 24

Graph $$\displaystyle{ \frac{(x-2)^2}{4} - \frac{(y+3)^2}{9} = 1 }$$.

solution

Practice 25

Graph $$\displaystyle{ \frac{(y+4)^2}{4} - \frac{(x-2)^2}{16} = 1 }$$.

solution

Practice 26

Graph $$4x^2-y^2=16$$.

solution

Practice 27

Graph $$-x^2+4y^2-2x-16y+11=0$$.

solution