<center> <img src = "https://i.imgsafe.org/d0/d09ff1a38f.png" height = "300" width = "400"/></center>
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# Parabola - Interesting Properties and Usages


## 1. From Cone to Parabola 
Conic section is a curve obtained as the intersection of the surface of a cone with a plane. If the cutting plane is parallel to exactly one generating line of the cone, then the conic is unbounded and is called a parabola. 

<center> <img src = "https://i.imgsafe.org/d1/d199decd7e.png" height = "400" width = "500"/></center>

Look at the following figure. 

<center> <img src = "https://i.imgsafe.org/d3/d3b7c7418f.png" height = "400" /></center>

1. Plane <img src="http://latex.codecogs.com/gif.latex?\alpha" title="\alpha" /> cross with the cone parallel to its generating line. 

2. Pick a point <img src="http://latex.codecogs.com/gif.latex?P" title="P" /> on the intersecting line between plane <img src="http://latex.codecogs.com/gif.latex?\alpha" title="\alpha" /> and the cone (which is painted in brown)

3. Then there exist unique sphere <img src="http://latex.codecogs.com/gif.latex?O" title="C" /> such that it contacts with the plane <img src="http://latex.codecogs.com/gif.latex?\alpha" title="\alpha" /> and the cone. 

<center> <img src = "https://i.imgsafe.org/d3/d3d01ccb0e.png" height = "400" /> </center>

Denote the point of contact between <img src="http://latex.codecogs.com/gif.latex?O" title="O" /> and <img src="http://latex.codecogs.com/gif.latex?\alpha" title="\alpha" /> as <img src="http://latex.codecogs.com/gif.latex?F" title="F" /> . 

4. Now, define the intersection between <img src="http://latex.codecogs.com/gif.latex?\overline{PA}" title="\overline{PA}" /> and <img src="http://latex.codecogs.com/gif.latex?O" title="O" /> as <img src="http://latex.codecogs.com/gif.latex?E" title="E" /> . Since <img src="http://latex.codecogs.com/gif.latex?\overline{PA}" title="\overline{PA}" /> and <img src="http://latex.codecogs.com/gif.latex?O" title="O" /> only intersects (i.e. in contact) in <img src="http://latex.codecogs.com/gif.latex?E" title="E" />, we get 

<center> <img src="http://latex.codecogs.com/gif.latex?\overline{PF}&space;=&space;\overline{PE}" title="\overline{PF} = \overline{PE}" /> </center> 

5. Line <img src="http://latex.codecogs.com/gif.latex?\ell&space;=&space;\alpha&space;\cap&space;\beta" title="\ell = \alpha \cap \beta" align = "center"/> where <img src="http://latex.codecogs.com/gif.latex?\beta" title="\beta" align = "center"/> is the plane perpendicular to the axis of rotation of the cone which has point <img src="http://latex.codecogs.com/gif.latex?E" title="E" /> in it. Define <img src="http://latex.codecogs.com/gif.latex?I" title="I" /> to be the foot of perpendicular from <img src="http://latex.codecogs.com/gif.latex?P" title="P" /> to <img src="http://latex.codecogs.com/gif.latex?\ell" title="\ell" /> . Then we get a nice property 

<center> <img src="http://latex.codecogs.com/gif.latex?\overline{PI}&space;=&space;\overline{PE}" title="\overline{PI} = \overline{PE}" /> </center>

so that <img src="http://latex.codecogs.com/gif.latex?\overline{PF}&space;=&space;\overline{PI}" title="\overline{PF} = \overline{PI}" /> . 

This means that the intersecting curve between the cone and plane <img src="http://latex.codecogs.com/gif.latex?\alpha" title="\alpha" /> is a parabola having focal point <img src="http://latex.codecogs.com/gif.latex?F" title="F" /> and directrix <img src="http://latex.codecogs.com/gif.latex?\ell" title="\ell" /> . 

## 2. Examples and Usages 

### Example 1. In physics
Consider the following cylindrical container with liquid to be rotating at uniform angualr velocity <img src="http://latex.codecogs.com/gif.latex?\omega" title="\omega" /> . 

<center> <img src = "https://i.imgsafe.org/d4/d41ed69a77.png" /> </center> [2]

The width of the container (or the diameter) is <img src="http://latex.codecogs.com/gif.latex?D" title="D" /> . Consider an infinitesimal liquid element <img src="http://latex.codecogs.com/gif.latex?dm" title="dm" /> at height <img src="http://latex.codecogs.com/gif.latex?h" title="h" /> above the minimum of the parabola. The forces acting on it are 

1. Gravitational force  <img src="http://latex.codecogs.com/gif.latex?\mathbf{g}dm" title="\mathbf{g}dm" align = "center"/>

2. Centripetal force <img src="http://latex.codecogs.com/gif.latex?dF_c&space;=&space;r\omega^2&space;dm" title="dF_c = r\omega^2 dm" align = "center"/> . 

Now the angle between the two forces <img src="http://latex.codecogs.com/gif.latex?\alpha" title="\alpha" /> will have the tangent value 

<center> <img src="http://latex.codecogs.com/gif.latex?\tan&space;\alpha=&space;\frac{dh}{dr}&space;=&space;\frac{dF_c}{\mathbf{g}dm}&space;=&space;\frac{r\omega^2}{\mathbf{g}}" title="\tan \alpha= \frac{dh}{dr} = \frac{dF_c}{\mathbf{g}dm} = \frac{r\omega^2}{\mathbf{g}}" /> </center>

This means that 

<center> <img src="http://latex.codecogs.com/gif.latex?\mathbf{g}dh&space;=&space;r\omega^2&space;dr" title="\mathbf{g}dh = r\omega^2 dr" /> </center>

Integrating both sides, 

<center> <img src="http://latex.codecogs.com/gif.latex?\int_0^{r}&space;r'\omega^2&space;dr'&space;=&space;\int_0^{h}&space;\mathbf{g}dh'&space;\implies&space;h&space;=&space;\frac{\omega^2&space;r^2}{2\mathbf{g}}" title="\int_0^{r} r'\omega^2 dr' = \int_0^{h} \mathbf{g}dh' \implies h = \frac{\omega^2 r^2}{2\mathbf{g}}" />  </center> which is the parabola. By rotation, the surface of spinning water would be the paraboloid, 

<center> <img src = "https://i.imgsafe.org/d4/d4557ca37e.png" height = "400" /></center>

### Example 2. Paraboloid mirror


<center> <img src = "https://i.imgsafe.org/d4/d4847c2d2a.png" height = "400"/> </center>

Suppose a light is reflected by the parabolic mirror at point <img src="http://latex.codecogs.com/gif.latex?P(x_0,&space;y_0)" title="P(x_0, y_0)" align = "center"/> , where the initial path of light was <img src="http://latex.codecogs.com/gif.latex?y&space;=&space;y_0" title="y = y_0" align = "center"/> . The Law of Reflection states that the angle between the incidence ray and mirror is equal to the angle between the reflected ray and mirror. 

Angle between <img src="http://latex.codecogs.com/gif.latex?\overline{PE}" title="\overline{PE}" /> and <img src="http://latex.codecogs.com/gif.latex?\ell" title="\ell" />: <img src="http://latex.codecogs.com/gif.latex?\theta_i" title="\theta_i" align = "center"/>

Angle between <img src="http://latex.codecogs.com/gif.latex?\overline{PF}" title="\overline{PF}" /> and <img src="http://latex.codecogs.com/gif.latex?\ell" title="\ell" />: <img src="http://latex.codecogs.com/gif.latex?\theta_o" title="\theta_o" />

Using dot product between directional vectors, 

<center> <img src="http://latex.codecogs.com/gif.latex?|\cos&space;\theta_i&space;|=&space;\frac{|(2p/y_0&space;,&space;-1)&space;\cdot&space;(0,&space;1)|}{\sqrt{(2p/y_0)^2&space;&plus;&space;1}\sqrt{1}}&space;=&space;\frac{1}{\sqrt{1&space;&plus;&space;(2p/y_0)^2}}" title="\cos \theta_i = \frac{(2p/y_0 , -1) \cdot (0, 1)}{\sqrt{(2p/y_0)^2 + 1}\sqrt{1}} = \frac{-1}{\sqrt{1 + (2p/y_0)^2}}" /> </center>

<center> <img src="http://latex.codecogs.com/gif.latex?|\cos&space;\theta_o|&space;=&space;\frac{|(y_0&space;,&space;p-x_0)&space;\cdot&space;(2p/y_0,&space;-1)|}{\sqrt{y_0^2&space;&plus;&space;(p-x_0)^2}\sqrt{1&space;&plus;&space;(2p/y_0)^2}}&space;=&space;\frac{|p&plus;&space;x_0|}{|p&space;&plus;&space;x_0|\sqrt{1&space;&plus;&space;(2p/y_0)^2}}&space;=&space;\frac{1}{\sqrt{1&space;&plus;&space;(2p/y_0)^2}}" title="|\cos \theta_o| = \frac{|(y_0 , p-x_0) \cdot (2p/y_0, -1)|}{\sqrt{y_0^2 + (p-x_0)^2}\sqrt{1 + (2p/y_0)^2}} = \frac{|p+ x_0|}{|p + x_0|\sqrt{1 + (2p/y_0)^2}} = \frac{1}{\sqrt{1 + (2p/y_0)^2}}" /> </center>

(using ) <img src="http://latex.codecogs.com/gif.latex?y_0^2&space;=&space;4px_0" title="y_0^2 = 4px_0" align = "center" />.

Now we get 

<center> <img src="http://latex.codecogs.com/gif.latex?\cos\theta_i&space;=&space;\cos&space;\theta_o&space;\implies&space;\theta_i&space;=&space;\theta_0" title="\cos\theta_i = \cos \theta_o \implies \theta_i = \theta_0" /> </center>

since all angles are accute. This means that every ray coming parallel with the axis of rotation will be concentrated to the focal point of parabola <img src="http://latex.codecogs.com/gif.latex?F" title="F" /> . 

1. We can apply this fact in making parabolic antennas, an antenna that uses a parabolic reflector, a curved surface with the cross-sectional shape of a parabola, to direct the radio waves. 

<center> <img src = "https://i.imgsafe.org/d4/d4bd584666.jpeg" width = "400"/></center> [3]

2. Also in solar power generation, the solar dish in parabolic shape concentrates the solar energy to its focal point, which greatly increases the heat generation. 

<center> <img src = "https://i.imgsafe.org/d4/d4c1fb0a79.png" width = "400"/> </center> [4]

### 3. Citation
[1] https://en.wikipedia.org/wiki/Conic_section (image is used)

[2] https://physics.stackexchange.com/questions/293106/shape-of-water-in-rotating-bucket (image is used)

[3] https://en.wikipedia.org/wiki/Parabolic_antenna (only image is used)

[4] https://www.researchgate.net/figure/Solar-dish-collector-with-conical-receiver-photographed-by-the-authors_fig2_269575959

All the other graphic images are made by myself using GeoGebra 3D plotter and Graphic Calculator.