1
$\begingroup$

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 :)

  • 0
    So, is this a question about mathematics, or about programming?2012-09-18
  • 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 $y=\pm \sqrt{\frac{1}{2}}$.2012-09-18
  • 0
    @GerryMyerson Mathematics ;)2012-09-18
  • 0
    @Cristian thanks... thats just too straight-forward ! @_@2012-09-18

1 Answers 1

2

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 $y=x=\pm \sqrt{\frac{3}{5}}$.

  • 0
    if a was 0 as well though, x could be basically anything.. right? how to handle that case?2012-09-18
  • 0
    @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 $x^2+4y^2=3$.2012-09-18
  • 0
    @foaly I added an edit to my answer...2012-09-18
  • 0
    Why in particular for x=y when it's for any x,y satisfying that 2nd equation?2012-09-18
  • 0
    @foaly it is true for any $x$ and $y$, so it must be true for $y=x$.2012-09-18
  • 0
    gotta go to class now. coming back to this later :P2012-09-18
  • 0
    Why do you ignore the solutions with $\: a = 0 = b \:$ and $\: x\neq y \:$? $\;\;$2012-09-18
  • 0
    @RickyDemer I did not ignore those. I only gave a particular one such that $x=y$.2012-09-18
  • 0
    I don't see the use of that... Since there are intersections for any x,y that satisfy the 2nd equation, so I'd say $y=\frac{\sqrt{3-x^2}}{2}$?2012-09-18
  • 0
    @RickyDemer what do you think?2012-09-18