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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

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) + 2x_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

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  • $\begingroup$ 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. $\endgroup$ – wychmaster Jul 8 at 6:17
  • $\begingroup$ I'll send you what i archieved today later cause i am busy with some other things, thank you for helping. $\endgroup$ – MrJab Jul 8 at 6:30
  • $\begingroup$ 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.. $\endgroup$ – MrJab Jul 8 at 9:16
  • $\begingroup$ I added the photo i got in code $\endgroup$ – MrJab Jul 8 at 9:17

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