Quick questions --> conics/graphing (1 Viewer)

[blackops23]

Hi guys quick question here:

We have the graph x^2 = 4 - 2y^2

--> obviously it is an ELLIPSE

but what if we take the positive square root the of the graph?

i.e. x = (4-2y^2)^0.5 --> how will this graph look like?

I had previously thought the graph x = (4-2y^2)^0.5 would have been the TOP HALF of the ellipse. But on graphmatica, I sketched x=(4-y^2)^0.5 and it showed a FULL ELLIPSE, the exact same graph as x^2 = 4 - y^2

But my booklet says that x=(4-2y^2)^0.5 is "HALF AN ELLIPSE"

That being said which one is right? And how would the negative square root look like? I.e. x = -(4-2y^2)^0.5 (on graphmatica, all 3 of THESE GRAPHS, were identical

--------------------

So guys please tell me, how would the negative and positive square roots of x^2 = 4-2y^2 differ from the original?

This seems like a really basic 2U question, but my head ain't working right now... :\

Thanks, appreciate the help!

 

[Hermes1]

thats not even an ellipse its a circle lol.

x^2 + y^2 = 4

circle centre 0,0 and radius 2

however x = sqrt(4-y^2) is a half ellipse

 

[Hermes1]

to draw sqrt(4-y^2)

first draw x = 4-y^2 which is just a parabola with the vertex on the x-axis and is concave down.
then just take the positive square root of that. what will happen is the vertex will move from (4,0) to (2,0) and it will kind of flatten out.

you then reflect this round the x-axis to find the negative square root bcuz it is an even function which means it has symmetry.

 

[blackops23]

AHHHH SHIT SHIT , i meant x=(4-2y^2)^0.5


guys can u please re do my question? i forgot the fucking 2

 

[AAEldar]

The graph for that is just the top half of the ellipse. It can't be the bottem half because you've taken the positive case. If you take the negative case it will be, and if you take both cases it will be the full ellipse.

http://www.wolframalpha.com/input/?i=x%3D%284-2y^2%29^0.5

 

[Hermes1]

it is still the same method i mentioned above and u still get a half ellipse for x = (4-2y^2)^0.5

all that changes is that the y int in x = (4-y^2)^0.5 was 2 and -2 but with the new graph it is y = +sqrt(2) and -sqrt(2).
but u still get the sam flatenning out and the same elliptical shape.

okay ill try and explain this another way. do u no how we hav y^2 = f(x) graphs in the curve sketching topic. well you no how we change that to y = +/- sqrt(f(x)) to draw it. thats the same thing we are doing with this graph.

u rearrange to get x^2 = 4-2y^2 and you change it to x = +/- sqrt(4-2y^2) and then draw it.

the only difference is the graph is of the form x = f(y).

 

[kooliskool]

I can't believe this lols, this is really a simple question, but noone corrected it...... the x=+square root… part means it's half an ellipse, but it's the right hand half, since x is greater than 0. If it's x=- square root…, it's the left hand half. It's actually the same reason as how you say the upper or lower half of a semicircle when it's y=…

 

[AAEldar]

The graph for that is just the top half of the ellipse. It can't be the bottem half because you've taken the positive case. If you take the negative case it will be, and if you take both cases it will be the full ellipse.

http://www.wolframalpha.com/input/?i=x%3D%284-2y^2%29^0.5

I can't believe this lols, this is really a simple question, but noone corrected it...... the x=+square root… part means it's half an ellipse, but it's the right hand half, since x is greater than 0. If it's x=- square root…, it's the left hand half. It's actually the same reason as how you say the upper or lower half of a semicircle when it's y=…

?

 

[Hermes1]

I can't believe this lols, this is really a simple question, but noone corrected it...... the x=+square root… part means it's half an ellipse, but it's the right hand half, since x is greater than 0. If it's x=- square root…, it's the left hand half. It's actually the same reason as how you say the upper or lower half of a semicircle when it's y=…

well tnk u mr obvious. if u did wat id said in my post u wuld have got this.

 

[cutemouse]

The graph for that is just the top half of the ellipse. It can't be the bottem half because you've taken the positive case. If you take the negative case it will be, and if you take both cases it will be the full ellipse.

http://www.wolframalpha.com/input/?i=x%3D%284-2y^2%29^0.5

It's the RHS half, because x>=0 (NB, y can still be < 0)

 

[Hermes1]

It's the RHS half, because x>=0 (NB, y can still be < 0)

thats what AAEldar is saying anyway. he saying the top half of the ellipse with respect to the y-axis.