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

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zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Let'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
 3 years ago
Best ResponseYou've already chosen the best response.2Doing 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
 3 years ago
Best ResponseYou've already chosen the best response.2Hmm lemme throw this equation into a grapher real quick. I wanna make sure I didn't miss anything.

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0thank you, very well put explanation ... can you graph these with a ti83 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
 3 years ago
Best ResponseYou've already chosen the best response.2\[\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=2\]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
 3 years ago
Best ResponseYou've already chosen the best response.2Calculator? :o hmm thinking.

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2If 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,\[\large y=\pm 2\sqrt{12x^2}\]And you'd probably have to put it in as 2 separate functions,\[\large y_1=2\sqrt{12x^2} \qquad \qquad y_2=2\sqrt{12x^2}\]

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2There's probably another way to do it, but I'm not too familiar with calculators :\

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.2If 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
 3 years ago
Best ResponseYou've already chosen the best response.2We 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.

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0these problems ask me to "graph several members of each family", and i assume they mean as C changes right

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0like just pick random C's and graph ??

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2When 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
 3 years ago
Best ResponseYou've already chosen the best response.2Oh it's not just growing faster, it's also moving the yintercept :O I didn't think about that.

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0does the members also include the original equation do you think?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2The equation we're graphing `is` the original equation, we've just moved some things around so it's easier to graph.

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0For 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.

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0or what "several members of each family" actually means

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2I'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
 3 years ago
Best ResponseYou've already chosen the best response.2I guessing that's probably what they want you to do, to graph a few of those.

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0That 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
 3 years ago
Best ResponseYou've already chosen the best response.2Oh I didn't notice that! Yah they intersect at right angles! Hah! That's cool :D

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0can anyone use that calculator... thats just on a website for everyone to use?

zepdrix
 3 years ago
Best ResponseYou've already chosen the best response.2Yah 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.

zonazoo
 3 years ago
Best ResponseYou've already chosen the best response.0well thank you again. for all your help.
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