Question

Semi-major axis length and semi-minor axis length of (x-2)%2 + 16(y-4)^2 = 4

Answer 1

Ellipse?

Answer 2

Yeah :D

Answer 3

Well, first thing you wanna do with ellipses is arrange them in this fashion...
\[\huge \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]Can you do that? We just need the right side to be equal to 1, which it isn't, currently...

Answer 4

Yeah I did it but I'm not if it's right hehe

Answer 5

Show me what you did, then ;)

Answer 6

\[\frac{ (x-2)^{2} }{ 4 } + \frac{ (y-4)^{2} }{ \frac{ 1 }{ 4 } } = 1\]

Answer 7

Gosh that was hard getting that right on here haha

Answer 8

You... have been holding out on me.
You're correct :)

Answer 9

Yay!

Answer 10

Okay just needed to check :p

Answer 11

I have another question... can I type it here?

Answer 12

Now, once you've gotten it in this form
\[\huge \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]
You need to find a and b, which should be easy enough?

Answer 13

Wait, we're done with this question? LOL What are your answers?

Answer 14

Yeah I'm done haha, semi-major is 2, semi-minor is 1/2 ?

Answer 15

Nicely done :)
Sure, put up your next question~

Answer 16

Thanks!

Answer 17

Thanks for the medal :D

Answer 18

Find the Cartesian equation of \[\left| z-2i \right| = \sqrt{2}\left| z-1 \right|\]

Answer 19

That was quite a jump. Hang on.

Answer 20

Haha yeah... okay thanks :D

Answer 21

What does Cartesian equation even mean? :/

Answer 22

z = x + yi

Answer 23

I see. Well, first, I'd think you should substitute x+yi for z...

Answer 24

Note: This is the first time I've heard of this, so go easy :D

Answer 25

Yeah I kinda did it, but it got really confusing and I couldn't continue haha

Answer 26

Oh okay haha

Answer 27

Well, if we substitute, we should get
\[\Large |x+yi-2i|=\sqrt2|x+yi-1|\]
right?

Answer 28

Isn't \[|z-2i|=\sqrt{(z-2i)(z+2i)}\]

Answer 29

It's complex numbers so I did |x + (y-2)| = root2 |(x-1) + y|

Answer 30

Okay wait

Answer 31

@agostino
it sure looks like it, but z itself is complex. Best separate it into real and imaginary components first.
\[\Large |x+(y-2)i|=|(x-1)+yi|\]
This is when it's rearragned, @stephanieee._

Answer 32

|z - 2i| is \[\sqrt{x ^{2} + (y-2)^{2}}\]

Answer 33

Yeah I got that :D

Answer 34

That it is, @stephanieee._
Are you sure you needed help with this? :/

Answer 35

Yeah but they want it in cartesian :(

Answer 36

So I have to expand and simplify I think

Answer 37

That shouldn't be too hard? :)

Answer 38

So you have
\[\Large \sqrt{x^2+(y-2)^2}=\sqrt{(x-1)^2+y^2}\]

Answer 39

The answer I got was \[-\frac{ 40 }{ 9 } = (x-2)^{2} + (y-\frac{ 2 }{ 3 })^{2} \] and that does not look right, I think I'm meant to have 1 on the right hand side

Answer 40

Yikes... I forgot the square root of 2 bit...
\[\Large \sqrt{x^2+(y-2)^2}=\sqrt2\sqrt{(x-1)^2+y^2}\]

Answer 41

Yep and then square both sides?

Answer 42

So, we square both sides...yes
\[\large x^2+(y-2)^2=2(x-1)^2+2y^2\]

Answer 43

And let the world tremble...
\[\large x^2+y^2-4y+4=2x^2-4x+2+2y^2\]

Answer 44

As it sees all you are about to accomplish...
\[\Large 2=x^2-4x+y^2-4y\]

Answer 45

Oh wait I think I got it now, I stuffed up on one of the calcs! LOL

Answer 46

It still leads to something negative on the left side, though... I think. Is that okay?

Answer 47

@terenzreignz since you started with complex numbers, i think that's fine.

Answer 48

Should be fine, I have to go eat now anyway :D

Answer 49

Thank you so much!

Answer 50

That's reassuring. I was kinda hoping it'd lead to an equation of an ellipse, though.

Answer 51

Thanks again :)

Answer 52

Oh but weren't you really finding the 'magnitude' of the complex numbers? Which is just a scalar isn't it...

Answer 53

I know... it's all perplexing
z = x+yi where x and y are real... so why did this happen?

Answer 54

Must have screwed up with the algebra bit.

Answer 55

well i guess x and/or y can still be negative...

Answer 56

Haha I think I just have to change to cartesian form :/

Answer 57

I got my wish (ellipse) after all, and no ordinary ellipse, but a circle :D

Answer 58

Yay thanks!! :D

Answer 59

@stephanieee._
WAIT
I DID screw up some algebra. Hold your horses...

Answer 60

\[\Large \sqrt{x^2+(y-2)^2}=\sqrt2\sqrt{(x-1)^2+y^2}\]
Squaring both sides
\[\Large {x^2+(y-2)^2}=2[(x-1)^2+y^2]\]
Simplifying
\[\Large x^2+y^2-4y+4 = 2[x^2-2x+1+y^2]\]\[\Large x^2+y^2-4y+4 = 2x^2-4x+2+2y^2\]
\[\Large 4-2=2x^2-x^2-4x+2y^2-y^2+4y\]\[\Large 2=x^2-4x+y^2+4y\]

Answer 61

^yep, i just found that. It looks like it'll change the center of the circle.
\[\Large 2=x^2-4x+y^2+4y \]\[\large 2 +4+4 = (x-2)^2 + (y+2)^2 \] \[\large 10 = (x-2)^2 + (y+2)^2 \]

Answer 62

Still a circle though
@stephanieee._
Please tell me you saw this -.-

Answer 63

Yeah I saw it haha

Answer 64

Just one little plus sign. If you were writing it, just one vertical slash through the minus sign in the "y-2" bit ;)

Good hunting :D

Answer 65

Hahah okay :)