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So we have an equation $x^2 + By^2=z^2$, and we're supposed to find for what values of $B$ this graph would show two intersecting planes. That value is $0$. Now I need to find the equation of these two planes. I'm not really sure how to start this one. I'm very shakey on this stuff. Any hints?

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up vote 3 down vote accepted

You write:

$$ x^2=z^2 $$ $$ x^2-z^2=0 $$ $$ (x+z)(x-z)=0 $$

that is satisfed by $z+x=0$ and $z-x=0$ that are you planes.

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