Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

we're supposed to write a small program that calculates the intersections of $a(x-1)-by=0$ and $x^2+4y^2=3$. So far that's not such a big deal. I successfully calculate both points for all $b \ne 0$ (since $0=(4\frac{a^2}{b^2}+1)x^2-8\frac{a^2}{b^2}x+4\frac{a^2}{b^2}-3$). But we're also supposed to give the intersection(s) for $b=0$, which is what I have no idea of.

Thanks for any help :)

share|improve this question
    
So, is this a question about mathematics, or about programming? –  Gerry Myerson Sep 18 '12 at 1:24
1  
if $b=0$, then $a(x-1)=0$ from the 1st equation. Hence, $x=1$ no matter what $a$ is. Then, the 2nd eq becomes $4y^2=2$ from where you get $y=\pm \sqrt{\frac{1}{2}}$. –  Cristian Sep 18 '12 at 1:29
    
@GerryMyerson Mathematics ;) –  foaly Sep 18 '12 at 1:37
    
@Cristian thanks... thats just too straight-forward ! @_@ –  foaly Sep 18 '12 at 1:38
add comment

1 Answer 1

up vote 2 down vote accepted

Perhaps, instead of comment I should've written it as my answer.

EDIT:

Suppose $a\neq 0$. If $b=0$, then $a(x−1)=0$ from the 1st equation. Hence, $x=1$ no matter what $a$ is. Then, the 2nd eq becomes $4y^2=2$ from where you get $y=\pm \sqrt{\frac{1}{2}}$.

If $a=0$ and $b=0$, then $a(x-1)-by=0$ for all $x\in \mathbb{R}$. In particular, it is true for $x=y$. Then, you get $y=x=\pm \sqrt{\frac{3}{5}}$.

share|improve this answer
    
if a was 0 as well though, x could be basically anything.. right? how to handle that case? –  foaly Sep 18 '12 at 1:42
    
@foaly. If $a$ and $b$ are both zero, your first equation could be satisfied by any $(x,y)$, so the intersection would then be any $(x,y)$ that satisfies $x^2+4y^2=3$. –  Rick Decker Sep 18 '12 at 1:44
    
@foaly I added an edit to my answer... –  Cristian Sep 18 '12 at 1:49
    
Why in particular for x=y when it's for any x,y satisfying that 2nd equation? –  foaly Sep 18 '12 at 1:54
    
@foaly it is true for any $x$ and $y$, so it must be true for $y=x$. –  Cristian Sep 18 '12 at 1:57
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.