complex # conic (1 Viewer)

[hatty]

sup again

|z - 2 | + | z + 2 | = 6

can someone find the cartesian equation of this 4 me plz and show me how it is an ellipse.


thanks alot.

 

[Calculon]

its an ellipse because it is a locus two focii at (2,0) and (-2,0) and the sum of the distance from these two focii are equal for the whole graph. As for that other bit, how about you try using the distance formula and add the distances from each focus to point P(x,y) and say that the sum of the two distances is equal to six. I haven't done conics yet so I might just be talking shit, I dunno.

 

[CM_Tutor]

In conics, it is easy to prove that for an ellipse with foci at S and S', PS + PS' = 2a. This equation is of that form, so this is an ellipse with (as Calculon noted), foci at z = 2 and z = -2.

We know 2a = 6, so a = 3.

To find b, consider what happens when z = bi, where b is real and positive. In this case, PS and PS' are equal (as triangles OPS and OPS' are congruent), so PS = 3 ______ (*).
But, PS^2 = 2^2 + b^2 = 4 + b^2 = 9 by (*)
So, b = sqrt(5), as b > 0.

Thus, the cartesian equation is x^2 / 9 + y^2 / 5 = 1.

Alternately, you could set z = x + iy and do a masive, aweful, algebra bash.

 

[hatty]

thanks alot guys

1 last question

it says

find the locus of z if

w = z-i/z-2 ( w is imaginary)




can u do a step by step walkthrough of this plz

btw
CM, r u a tutor?

 

[hatty]

and

w = z-2 / z + 2

arg w = pi/3

 

[Calculon]

Hes like got a bachelor of something and getting a doctorate in teaching, or something

 

[Faera]

With the "w = z-i/z-2 ( w is imaginary)" question, let z = x + iy
Group all the real, imaginary terms of both the numerator and denominator.
Then realise the denominator (by multiplying top and bottom by the conjugate demonimator).
Then regroup the real/imaginary terms of the numerator, and let Re(w) = 0 (since w is imaginary)
Don't worry about anything you've got in the denominator- when you multiply it up, it disappears (because of the 0).
Then you've got the locus of w! (in terms of x, y)

For the second one,
if argw = pi/3
arg((z-2)/(z+2)) = pi/3
that is, arg(z-2) - arg(z+2) = pi/3
Vector from z to (2,0) and vector from z to(-2,0) intersect at point z, and the angle between them is pi/3 - it's the major arc of the circle...
i think.

 

[CM_Tutor]

Originally posted by hatty
btw
CM, r u a tutor?

Yes, I do some tutoring, but I'm really a graduate student at Uni, doing a PhD in Chemistry Education and also an MEd in Teaching and Curriculum Studies. As Calculon mentioned, I have a BSc(Hons), too.

 

[spice girl]

Originally posted by CM_Tutor
Yes, I do some tutoring, but I'm really a graduate student at Uni, doing a PhD in Chemistry Education and also an MEd in Teaching and Curriculum Studies. As Calculon mentioned, I have a BSc(Hons), too.

hey CM, wot's ur thesis on, can u tell?

and are u planning to teach high school chem in the future? or something higher?

 

[CM_Tutor]

Originally posted by spice girl
hey CM, wot's ur thesis on, can u tell?

Looking at correlations between perceived understanding by 1st year Chem students and performance as measured in assessments.

and are u planning to teach high school chem in the future? or something higher?

I doubt it - amazingly I won't be qualified to teach in a school, according to the department of education, even once I have graduated. Becoming an academic is a possibility - as you have no doubt noticed, some of our Uni teaching could be better, so it might be nice to work on that. There are a few other possibilities, too.