## 17Calculus - Parabolas in Rectangular Coordinates

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A parabola is formed when a plane intersects a cone parallel to the side of the cone. This intersection produces only one curve (as compared to two in a hyperbola). On this page, we discuss parabolas in rectangular coordinates. For a discussion of parabolas in polar form, see this separate page.

Parabola

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 Figure 2.

A special comment is in order. 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 this a parabola. In words this says, for every point on the parabola, the distance between the point and the focus is the same as the distance from the point and the perpendicular distance to the directrix.

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

Figure 3 [Built with GeoGebra]

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

vertex

$$(h,k)$$

axis of symmetry

$$y=k$$

focus

$$(h+p,k)$$

directrix

$$x=h-p$$

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

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

Figure 4 [Built with GeoGebra]

Practice

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

Problem Statement

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

Solution

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

Problem Statement

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

Solution

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

Problem Statement

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

Solution

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

Problem Statement

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

Solution

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### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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