Hi, i got this question and dont fully understand how they solved it:
You have to find the locus of points representing z if:
abs(z-3i)-abs(z+31)=2
my first thought would be to replace z with (x+iy) but what then?
Hi, i got this question and dont fully understand how they solved it:
You have to find the locus of points representing z if:
abs(z-3i)-abs(z+31)=2
my first thought would be to replace z with (x+iy) but what then?
Then take the absolute value of x + iy - 3i = x + i(y - 3) and x + iy + 3i = x + i(y + 3).
So you have to simplify something of the form $\displaystyle \sqrt{A} - \sqrt{B} = 2 \Rightarrow \sqrt{A} = 2 + \sqrt{B}$.
Square both sides, simplify, square both sides again and simplify. You will get a single branch of a hyperbola (there is an implicit restriction on z you will need to be careful of that. This restriction means that both branches do no get included).
Of course, if you're familiar with the locus definition of a hyperbola then there's a much simpler geometric approach that can be taken.
Okay, seems quite simple now. Thanks. This is what i have done:
$\displaystyle \sqrt{x^2+(y-3)^2}-\sqrt{x^2+(y+3)^2}=2$
$\displaystyle \sqrt{x^2+(y-3)^2}=2+\sqrt{x^2+(y+3)^2}$
$\displaystyle x^2+(y-3)^2=4+4\sqrt{x^2+(y+3)^2}+x^2+(y+3)^2$
$\displaystyle x^2+y^2-6y+9=4+4\sqrt{x^2+(y+3)^2}+x^2+y^2+6y+9$
$\displaystyle -12y-4=4\sqrt{x^2+(y+3)^2}$
$\displaystyle -(3y+1)=\sqrt{x^2+(y+3)^2}$
$\displaystyle (3y+1)^2=x^2+(y+3)^2$
$\displaystyle x^2-8y^2+8$
$\displaystyle x^2/8-y^2=1$
And in this case it will only be the bottom branch of the graph.
i have also looked at the locus definition of the hyperbola, didnt think of defining it that way. makes it much simpler just by substituting in:
$\displaystyle \sqrt{(x-c)^2+y^2}-\sqrt{(x+c)^2+y^2}=2a$
into:
$\displaystyle x^2/a^2-y^2/(c^2-a^2)=1$
you get the final equation.