<p>Find the formulas of the middle point algorithm for drawing the parabola ? <br>
x^2 = 4*(a+b)*y , a,b &gt; 0. <br>
Solve it according to E(East) and NorthEast(NE) ? <br></p>
<p>This is the result i got (This is writen in java frame, so its the opposite on graph)</p>
<p><a href="https://i.sstatic.net/OLfzY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OLfzY.png" alt="My try" /></a><br />
<br>
This is my try, Ok in this problem the y increases y_{k+1} = y_k + 1 every time and x_{k+1} = x_k or x_k + 1 depending on the P_k value. Here is code i tried with: If P_{k} &gt; then ( p_{k+1} = p_k - 4*(a + b) + 2<em>x_k + 1; and x in increased by one x_{k+1} = x_k + 1;) else ( p_{k+1} = p_k - 4</em>(a + b);) , It doesn't look very well i think</p>

CC BY-SA 4.0
2020-07-08T06:17:09.683Z
This is obviously a homework/exercise question. So can you tell us what you have tried by yourself and what you are struggling with? I searched the internet for "middle point algorithm parabola" and already got some promising results.  Just copy-pasting the question is at least for me a bit to less effort on your side to be willing to help.
wychmaster
CC BY-SA 4.0
2020-07-08T06:30:31.077Z
I'll send you what i archieved today later cause i am busy with some other things, thank you for helping.
MrJab
CC BY-SA 4.0
2020-07-08T09:16:26.620Z
https://www.youtube.com/watch?v=Y0Onvip-j9w ... By watching this video this is my try,  Ok in this problem the y increases y_{k+1} = y_k + 1 every time and x_{k+1} = x_k or x_{k+1} depending on the P_k value.  Here is code i tried with:  If P_{k} >   then  (   p_{k+1} = p_k - 4*(a + b) + 2*x_k + 1;  and x in increased by one  x_{k+1} = x_k + 1;)  else (  p_{k+1} = p_k - 4*(a + b);) , It doesn't look very well i think..
MrJab
CC BY-SA 4.0
2020-07-08T09:17:28.940Z
I added the photo i got in code
MrJab
<midpoint>
Find the formulas of the middle point algorithm for drawing the parabola?
MrJab
# Find the formulas of the middle point algorithm for drawing the parabola?

(Asked by: username_0 on 2020-07-08)

Find the formulas of the middle point algorithm for drawing the parabola ?   
x^2 = 4\*(a+b)\*y , a,b > 0.   
Solve it according to E(East) and NorthEast(NE) ?

This is the result i got (This is writen in java frame, so its the opposite on graph)

[![My try](https://i.sstatic.net/OLfzY.png)](https://i.sstatic.net/OLfzY.png)  
  
This is my try, Ok in this problem the y increases y\_{k+1} = y\_k + 1 every time and x\_{k+1} = x\_k or x\_k + 1 depending on the P\_k value. Here is code i tried with: If P\_{k} > then ( p\_{k+1} = p\_k - 4\*(a + b) + 2*x\_k + 1; and x in increased by one x\_{k+1} = x\_k + 1;) else ( p\_{k+1} = p\_k - 4*(a + b);) , It doesn't look very well i think

**Tags:** midpoint

> **Comments:**
> **username_1**: This is obviously a homework/exercise question. So can you tell us what you have tried by yourself and what you are struggling with? I searched the internet for "middle point algorithm parabola" and already got some promising results.  Just copy-pasting the question is at least for me a bit to less effort on your side to be willing to help.  
> **username_0**: I'll send you what i archieved today later cause i am busy with some other things, thank you for helping.  
> **username_0**: https://www.youtube.com/watch?v=Y0Onvip-j9w ... By watching this video this is my try,  Ok in this problem the y increases y_{k+1} = y_k + 1 every time and x_{k+1} = x_k or x_{k+1} depending on the P_k value.  Here is code i tried with:  If P_{k} >   then  (   p_{k+1} = p_k - 4*(a + b) + 2*x_k + 1;  and x in increased by one  x_{k+1} = x_k + 1;)  else (  p_{k+1} = p_k - 4*(a + b);) , It doesn't look very well i think..  
> **username_0**: I added the photo i got in code  

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## 0 answers

