How do we graph this ellipse?
2x^2 + (y^2)/2 = 1

- anonymous

How do we graph this ellipse?
2x^2 + (y^2)/2 = 1

- jamiebookeater

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- zepdrix

Let's get it into standard form.\[\large \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1\]
Hmm, so the x has a 2 on it.
Multiplying by 2 is the same as dividing by 1/2.\[\large 2x^2 \qquad \rightarrow \qquad \frac{x^2}{\left(\frac{1}{2}\right)}\]Then let's apply the square to both the top and bottom term.\[\large \frac{x^2}{\left(\frac{1}{2}\right)} \qquad \rightarrow \qquad \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2\]Now we can clearly see what our \(a\) value is.

- zepdrix

Doing similar to the y gives us something like this I believe,
\[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=1\]
And then to graph it.. ummm trying to remember.
If x=0, y extends out to \(\pm \sqrt2\).
if y=0, x extends out to \(\pm \dfrac{1}{\sqrt2}\)

- zepdrix

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## More answers

- zepdrix

Hmm lemme throw this equation into a grapher real quick. I wanna make sure I didn't miss anything.

- zepdrix

Yah it looks good.

- anonymous

thank you, very well put explanation ... can you graph these with a ti-83 or anything like that? and if that equation was set equal to 2 instead of 1, you would just set each variable = to zero and then solve for your new x and y

- zepdrix

\[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=2\]Divide both sides by 2,\[\large \left(\frac{x}{\frac{\sqrt2}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2\cdot \sqrt2}\right)^2=1\]Which simplifies to,\[\large \left(\frac{x}{1}\right)^2+\left(\frac{y}{2}\right)^2=1\]These would be our new \(a\) and \(b\) values.

- zepdrix

Calculator? :o hmm thinking.

- zepdrix

If you wanted to put it into your calculator, you would probably want to avoid the fancy ellipse standard form.
From the start of the problem, \[\large 2x^2 + (y^2)/2 = 1\]Solving for y gives us,\[\large y=\pm 2\sqrt{1-2x^2}\]And you'd probably have to put it in as 2 separate functions,\[\large y_1=2\sqrt{1-2x^2} \qquad \qquad y_2=-2\sqrt{1-2x^2}\]

- zepdrix

There's probably another way to do it, but I'm not too familiar with calculators :\

- anonymous

so we always want it in the form of =1 then...
the next problem i got was -x + ln y = C
how would we graph that as C changes?

- zepdrix

If we solved for y, we would see that this is an exponential function.\[\large \ln y=C+x\]Exponentiating both sides,\[\large e^{\ln y}=e^{C+x} \qquad \rightarrow \qquad y=e^Ce^x\]
As C changes? :O hmm it looks like it's just changing the scaling on our function. It just grows a lot faster as C increases.

- zepdrix

We can't graph an unknown constant :D I guess you could just graph a few different functions and compare.
With different coeeficients in front, e^2, e^3 and so on.

- anonymous

these problems ask me to "graph several members of each family", and i assume they mean as C changes right

- anonymous

like just pick random C's and graph ??

- zepdrix

When C=0, we have the graph \(y=e^x.\)
When C=1, we have the graph \(y=e\cdot e^x\), so this one will grow about 3 times as fast as e^x.
When C=2, we have, \(y=e^2\cdot e^x\), so this one will grow like 7 times faster than e^x.
Yah just plug in random C's. Graph a few members of the family! :)

- zepdrix

Oh it's not just growing faster, it's also moving the y-intercept :O I didn't think about that.

- zepdrix

https://www.desmos.com/calculator/2saf2toes8

- anonymous

does the members also include the original equation do you think?

- zepdrix

The equation we're graphing `is` the original equation, we've just moved some things around so it's easier to graph.

- zepdrix

https://www.desmos.com/calculator/agvkcwjylj

- anonymous

For example... the first one... the problem asks us to find the orthogonal trajectories for the family of curves and graph several members of each family.
x = Cy^4 was the original problem and after we find the orthogonal trajectories our equation becomes
2x^2 + y^2/2 = C
so I am unsure if I should just graph random C's of the resulting equation or both.

- anonymous

or what "several members of each family" actually means

- zepdrix

I'm not exactly sure about the orthogonal trajectories,
But here are several members of each family graphed,
The equations should be displayed on the left.
https://www.desmos.com/calculator/mqoozlfnq2

- zepdrix

I guessing that's probably what they want you to do, to graph a few of those.

- anonymous

That is a cool looking graphing calculator... but i see that both equations are orthogonal... so I guess we did the problem right... so I would just graph both then since they probably want to see them orthogonal

- zepdrix

Oh I didn't notice that!
Yah they intersect at right angles! Hah! That's cool :D

- anonymous

can anyone use that calculator... thats just on a website for everyone to use?

- zepdrix

Yah I can't remember who showed it to me.
Maybe Satellite or someone..
But I had to bookmark it immediately, very useful tool! :)
Yah it's just a free resource I think.

- anonymous

well thank you again. for all your help.

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