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?
Finding equation of a plane
0
$\begingroup$
calculus
1 Answers
3
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.