# intersections of two functions, division by zero

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 :)

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So, is this a question about mathematics, or about programming? –  Gerry Myerson Sep 18 '12 at 1:24
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

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 $y=x=\pm \sqrt{\frac{3}{5}}$.
@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 it is true for any $x$ and $y$, so it must be true for $y=x$. –  Cristian Sep 18 '12 at 1:57