- #1
autodidude
- 333
- 0
Find the locus defined by [tex]|z-2|-|z+2|=3[/tex]
The given example rewrites the left hand side as:
[tex]\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2}[/tex]
1) When they rewrite it as that, they square it and square root it right? Why isn't it squaring the whole expression and rooting it (isn't that the rule?)
i.e.
[tex]\sqrt{((x-2)+y)^2}-\sqrt{((x+2)+y)^2}[/tex]
2) Why does the 'i' disappear?
Then they move the second expression to the right hand side, square both sides and complete the square for bot sides, square both sides again and they end up with the locus being a hyperbola with asymptotes at y=±√(7)x/3 and x≤-(3/2)
They don't explain the reason for taking these steps (terse book) - is it just standard procedure for these type of questions? To simplify + complete the square until you get a familiar looking equation?
The given example rewrites the left hand side as:
[tex]\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2}[/tex]
1) When they rewrite it as that, they square it and square root it right? Why isn't it squaring the whole expression and rooting it (isn't that the rule?)
i.e.
[tex]\sqrt{((x-2)+y)^2}-\sqrt{((x+2)+y)^2}[/tex]
2) Why does the 'i' disappear?
Then they move the second expression to the right hand side, square both sides and complete the square for bot sides, square both sides again and they end up with the locus being a hyperbola with asymptotes at y=±√(7)x/3 and x≤-(3/2)
They don't explain the reason for taking these steps (terse book) - is it just standard procedure for these type of questions? To simplify + complete the square until you get a familiar looking equation?