Q 2 d) ellipse (2 Viewers)

[danni a]

with this question how did u do it... i wsnt sure cos ive only ever seen circles on the argand diagram not ellipses


|z-1-3i| + |z-9-3i| = 10

 

[joesmith1975]

i didnt do it but i when to my teacher and he said u remember the property of the ellispies and i got NO and then he goes PS + PM = 2a so the major is 5 and i got dam i guessed 6

 

[dongypro]

its just like any other locus

at first i was like wtf

but u can tell the ends of the ellispse is 1 and 9 with the same height 3

1+9 / 2 = 5

so centre for x value is 5

centre (5,3)

5 + 3i (argh i forgot to put i rofl)


i tried the long way also by letting z = x + iy

then squaring , and u get like this random equation and i found it was 5,3 too

come to think about it,, i think i mite've screwed this question up ... nOooooooooo

i dunno lol~ i dun wanna think about 4unit maths ever again...

 

[chucknthem]

I got liek a circle with center (5,3) and radius 5, lol. something went wrong

 

[zeek]

for part d i said that the range for arg(z) was where the vectors from the origin touched the endpoints of the ellipse and then gave a value for it.

i got something<=arg(z)<=pi/2

 

[gamecw]

omfg whats majar n minor again??

i got 10 n 6 centre (5,3)

r u supposed to write 5 n 3????????????
idid write 10(2x5)
6(2x3)

 

[pLuvia]

Centre was (5,3)

 

[jon0_o]

i spent agesss on this q saying z=x+iy .. but im really pissed cos i wasn't sure if the major axis was a or 2 a .. so i just put 10 . but was unsure for the tie i spent .. and for one part i didn't put it as a complex number :S .. i put 5,3 instead of 5+3i ... n the time there costs me lyk 6-7 marks later on

 

[zeek]

major axis is 2a semi-major is a
minor axis is 2b semi-minor is b

i used the eccentricity equation to find b.

 

[shsshs]

the centre was 5 + 3i, not (5,3)
it asked for the complex representation

also remember u when u draw the ellipse, it hasta 'touch' the x and y axis

 

[gamecw]

major axis is 2a semi-major is a
minor axis is 2b semi-minor is b

i used the eccentricity equation to find b.

THANK GOD

i just drew 2 lines to the mid (top) of ellipse.. then use a^2+b^2=c^2 to find the half minor, then times it by 2!

and same for the arg(z) at that pt, angle's a maximum, then minimum is negative that number

 

[Ror bones]

considering that d) iii)

"find range of arg(z) for z points on ellipse"

was only a 1 mark question it was a give away that the ellipse touches x and y otherwise it would be to hard

that way i figured out the major and minor axes and then checked it with a point that would lie on one of the axes

bhaha

 

[zeek]

actually... since one of the focii was 1+3i and x=ae then e=1/5
if you subbed this into b^2=a^2(1-e^2) and then find b, you will see that the ellipse cuts the x-axis so you had to calculate the next complex number and then find the argument from there.

 

[tangy_lilac888]

Aghhh gahh, i totally stuffed it XD.
Coudln't square properly and got a stupid circle.
Did it second time and STILL got a circle.
THen, because question called for ellipse, I managed to convince myself that a "circle" was a SPECIAL type of "ellipse".
Yikes.

 

[Riviet]

I did it the algebraic way as I wasn't sure how to work out the features of the ellipse using the starting equation.

Let z=x+iy.... etc.

 

[pLuvia]

I did it the algebraic way as I wasn't sure how to work out the features of the ellipse using the starting equation.

Let z=x+iy.... etc.

Same, took way too long

 

[jctaylor]

Probably the easiest way to do this question would be to recognise that the points in the question are the foci, then find that the locus has to pass thru 3i and 10 + 3i
which means the centre would be at 5 + 3i, so the max y has to have Im part of 3i

 

[Riviet]

Probably the easiest way to do this question would be to recognise that the points in the question are the foci, then find that the locus has to pass thru 3i and 10 + 3i
which means the centre would be at 5 + 3i, so the max y has to have Im part of 3i

I knew there was that geometric method, but wasn't sure of it so just decided to go algebraic.

 

[Shady01]

I went algebraically, then grouped them:
(x-1)^2 + (y-3)^2 + (x-9)^2 + (y-3)^2= 100 (thats how i was shown to get rid of absolute vaule thingy

then i expanded: x^2-2x+1+y^2-6y+9+x^2-18x+81+y^2-6y+9=100
: 2x^2-20x +2y^2-12y = 0
: div by 2 and complete square
: x^2-10x +25+y^2-6y+9=25+9
: (x-5)^2 + (y-3)^2= 34
so i thought ellipse was:
[(x-5)^2]/34 + [(y-3)^2]/34=1
so i said my centre was 5/34, 3/34?

what did i do wrong?

 

[Mill]

Shady01, your first line is wrong.

If you were to go down the z = x + iy path, using modulus definitions:

sqrt(c) + sqrt(d) = 10

You can't just square both sides to get c + d = 100.


In fact, the easiest way to do this is to rearrange as sqrt(c) = 10 - sqrt(d) and then square both sides.


Regarding the geometric method, as people said, you're supposed to know that is the locus of an ellipse and relate it to the equation PS + PS' = 2a.
Then you can find S(1,3), S'(9,3) and a = 5 really quickly.

Then, if you were to draw a diagram, you could find the value of b = 3 through some nifty Pythagoras or some other method.