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

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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.
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 \(\pm \dfrac{1}{\sqrt2}\)

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Hmm lemme throw this equation into a grapher real quick. I wanna make sure I didn't miss anything.
Yah it looks good.
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
\[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=2\]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.
Calculator? :o hmm thinking.
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,\[\large y_1=2\sqrt{1-2x^2} \qquad \qquad y_2=-2\sqrt{1-2x^2}\]
There's probably another way to do it, but I'm not too familiar with calculators :\
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?
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.
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.
these problems ask me to "graph several members of each family", and i assume they mean as C changes right
like just pick random C's and graph ??
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! :)
Oh it's not just growing faster, it's also moving the y-intercept :O I didn't think about that.
does the members also include the original equation do you think?
The equation we're graphing `is` the original equation, we've just moved some things around so it's easier to graph.
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.
or what "several members of each family" actually means
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.
I guessing that's probably what they want you to do, to graph a few of those.
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
Oh I didn't notice that! Yah they intersect at right angles! Hah! That's cool :D
can anyone use that calculator... thats just on a website for everyone to use?
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.
well thank you again. for all your help.

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