Complex numbers! (2 Viewers)

[Aysce]

Yo yo - Just need help with 1 or 2 generic questions i've forgotten to do (Yes - shame on me)




And part b of this question:



Thanks guise =)

 

[barbernator]

i will give u clues not solutions

e) 10 is the sum of the distances from each focus. So the foci are (0,-2) and (0,6). So now we need to work out the major and minor axes of the ellipse. Think geometrically in terms of a right angled triangle to find the minor axis length. To find the major axis length, use a simple geometric property of the ellipse to do with the distances from foci. And centre is obviously just half way between foci.

b) Smash out some algebra by subbing in the values for z1 and z2, and for z = x+iy. Group real and imaginary parts, so you can find the modulus of each, and then square both sides. Then a heap of cancelling should occur and it will be the equation of a circle.

 

[deswa1]

Before Barb posts

I think the first one is from 06 HSC? Anyway, if you remember properties of an ellipse, you know that the sum of the distances from the two foci equals 2a. So now you know the length of the major axis, the two foci, you can find the centre and then you should be able to go from there.

For the second one, plot the two points. Now look at the diagram and think what sort of line will be equidistant from both of those points all the time.

 

[deswa1]

Let me know if you want more detailed stuff- these were just meant to point in a general direction- Barb will probs post something better than mine anyway so it should be good

 

[SpiralFlex]

Teh Cicool of Appollinus

 

[Nooblet94]

Yo yo - Just need help with 1 or 2 generic questions i've forgotten to do (Yes - shame on me)




And part b of this question:



Thanks guise =)

An ellipse has the property PS+PS'=2a where PS and PS' are the distances from a point on the ellipse to the two focii and 2a is the length of the major axis. So the ellipse in question has focii (-2,0) and (6,0) and major axis 10. From there it's fairly straightforward. (Just so you're aware, the hyperbola has a similar property, except it's the difference in the distance to the focus)

a. will be the line that is the perpendicular bisector of z1 and z2
b. Just let z=x+iy, expand both sides and simplify. You can do a. this way too, it just takes longer.


(damn, beaten by 3 guaranteed state rankers )

 

[SpiralFlex]

Clue: On a complex plane, the meaning of |z-2| is ______________________ .

 

[SpiralFlex]

An ellipse has the property PS+PS'=2a where PS and PS' are the distances from a point on the ellipse to the two focii and 2a is the length of the major axis. So the ellipse in question has focii (-2,0) and (6,0) and major axis 10. From there it's fairly straightforward. (Just so you're aware, the hyperbola has a similar property, except it's the difference in the distance to the focus)

a. will be the line that is the perpendicular bisector of z1 and z2
b. Just let z=x+iy, expand both sides and simplify. You can do a. this way too, it just takes longer.

Do you ruin movies as well? :/

 

[Aysce]

Alright cheers boys, I'll try again =)

 

[SpiralFlex]

Aysce, the most important thing is you understand why.

 

[Nooblet94]

Do you ruin movies as well? :/

Yup

 

[Miss World]

Aysce, the most important thing is you understand why.

+1

 

[SpiralFlex]

A good way of testing understanding is to explain it to people.

 

[Miss World]

i will give u clues not solutions

e) 10 is the sum of the distances from each focus. So the foci are (0,-2) and (0,6). So now we need to work out the major and minor axes of the ellipse. Think geometrically in terms of a right angled triangle to find the minor axis length. To find the major axis length, use a simple geometric property of the ellipse to do with the distances from foci. And centre is obviously just half way between foci.

b) Smash out some algebra by subbing in the values for z1 and z2, and for z = x+iy. Group real and imaginary parts, so you can find the modulus of each, and then square both sides. Then a heap of cancelling should occur and it will be the equation of a circle.

U sure you didn't swap x and y coordinates ?

 

[Aysce]

U sure you didn't swap x and y coordinates ?

He meant (-2,0) and (6,0)

By the way guys, clocked it and understand how this shieeeet works now woop woop!

But I don't really understand how to do part B without using algebra? Should I just use algebra - I want a faster way

 

[Miss World]

He meant (-2,0) and (6,0)

By the way guys, clocked it and understand how this shieeeet works now woop woop!

But I don't really understand how to do part B without using algebra? Should I just use algebra - I want a faster way

fahk, that confused the shitt outta me!

and algebra doesnt take much time

 

[deswa1]

He meant (-2,0) and (6,0)

By the way guys, clocked it and understand how this shieeeet works now woop woop!

But I don't really understand how to do part B without using algebra? Should I just use algebra - I want a faster way

Algebra that one defs- there might be a way to do it geometrically but I don't know it. Regardless, the algebra isn't long really.

 

[SpiralFlex]

Generally the Circle of Apollonius is in the form of,








Can you think of a reason why lambda cannot be equal to one. (Clue, bisectors)

 

[seanieg89]

Generally the Circle of Apollonius is in the form of,








Can you think of a reason why lambda cannot be equal to one. (Clue, bisectors)

Pretty sure you mean not equal to 1 either. 0 gives a point and 1 gives a line, (which can be interpreted as a circle of infinite radius).

 

[SpiralFlex]

Pretty sure you mean not equal to 1 either. 0 gives a point and 1 gives a line, (which can be interpreted as a circle of infinite radius).

yeah sorry typo,