Thread: Finding intersection points of two ellipses (Using C program)

Today i am going find out solution of two ellipses intersection points using C programming but i am not find out know idea, I solved using geometry equation substitute method but i am not unable to do same thing in C programming please help me .I am talking example as following two ellipses
(x^2)/4+y^2=1 ,
((x-2)^2)/4+y^2=1
please give me any suggestion so i will proceed head.

What have you tried to do in C?

What grumpy was asking you to do is provide an effort.

We already know what you want.

Read the site's homework policy, here. This is not a free code factory. If you want help, demonstrate a genuine effort to solve your problem on your own - with code. Then people might consider helping you with specific problems you have encountered.

But, then again, from your behaviour, what you seek is someone to write the code for you, rather than doing it yourself. That won't happen.

Originally Posted by grumpy

We already know what you want.

Read the site's homework policy, here. This is not a free code factory. If you want help, demonstrate a genuine effort to solve your problem on your own - with code. Then people might consider helping you with specific problems you have encountered.

But, then again, from your behaviour, what you seek is someone to write the code for you, rather than doing it yourself. That won't happen.

Today i tried following is my code :

Code:

#include<stdio.h>
#include<math.h>


int intersection(float,float,float,float,float,float,float,float);
int main()
{
    int x1,y1,a1,b1,x2,y2,a2,b2;
    int ret;
//    ret=intersection(x1,y1,a1,b1,x2,y2,a2,b2);
    ret=intersection(2,2,2,1,7,2,2,1);
    if(ret==1)
        printf("Two ellipses are intersect\n");
    else
        printf("Two ellipses are not intersect\n");




    return 0;
    
}






int intersection(float x1,float y1,float a1,float b1,float x2,float y2,float a2,float b2)
{
    float x,y;
    float z;
    x=(1/(a2*b1-a1*b2)*(a2*b1-a1*b2))*(-sqrt(-(a1*a1*a2*a2)*(y1-y2)*(y1-y2)*(a2*b1-a1*b2)*(a2*b1-a1*b2)));
    y=(1/(a2*b1-a1*b2)*(a2*b1-a1*b2))*((a1*a1*b2*b2*x2)-(a1*a2*b1*b2*x1)-(a1*a2*b1*b2*x2)+(a2*a2*b1*b1*x1));
    z=(1/(a2*b1-a1*b2)*(a2*b1-a1*b2));
    printf("result of intersection is %f   and %f   and  %f",x,y,z);


    return 1;
}

after compiling it giving following output:
result of intersection is -nan and -nan and -nanTwo ellipses are intersect

I am getting -nan as output why i am not understand
please help me

Check to see if you are dividing by 0.

You don't really want an explanation of why you are getting NaNs, so I'm not even going to bother with that.

To fix the problem, try breaking down your calculations into smaller parts, and check each of the parts before combining them.

Before computing a/b, check that b (however you calculated it) is not zero. Before computing sqrt(something), check that the something is not negative.

Dividing by zero doesn't produce NaNs. However, multiplying the result of that (an infinity) by zero does produce NaNs.




Note: not all compilers support IEEE floating point format. If your code is built with such a compiler, it will not produce NaNs - instead you'd be asking questions about program crashes (e.g. reporting a division by zero, or some such).

Your function intersect currently returns 1 in all cases, and your main program interprets this to mean that they intersect in all cases. Maybe you forgot a step in your function? E.g. a conditional to say whether the ellipses intersect