## zonazoo Group Title How do we graph this ellipse? 2x^2 + (y^2)/2 = 1 one year ago one year ago

1. zepdrix Group Title

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.

2. zepdrix Group Title

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 $$\pm \dfrac{1}{\sqrt2}$$

3. zepdrix Group Title

|dw:1359930596041:dw|

4. zepdrix Group Title

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

5. zepdrix Group Title

Yah it looks good.

6. zonazoo Group Title

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

7. zepdrix Group Title

$\large \left(\frac{x}{\frac{1}{\sqrt2}}\right)^2+\left(\frac{y}{\sqrt2}\right)^2=2$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.

8. zepdrix Group Title

Calculator? :o hmm thinking.

9. zepdrix Group Title

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,$\large y_1=2\sqrt{1-2x^2} \qquad \qquad y_2=-2\sqrt{1-2x^2}$

10. zepdrix Group Title

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

11. zonazoo Group Title

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?

12. zepdrix Group Title

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.

13. zepdrix Group Title

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.

14. zonazoo Group Title

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

15. zonazoo Group Title

like just pick random C's and graph ??

16. zepdrix Group Title

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! :)

17. zepdrix Group Title

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

18. zepdrix Group Title
19. zonazoo Group Title

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

20. zepdrix Group Title

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

21. zepdrix Group Title
22. zonazoo Group Title

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.

23. zonazoo Group Title

or what "several members of each family" actually means

24. zepdrix Group Title

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

25. zepdrix Group Title

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

26. zonazoo Group Title

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

27. zepdrix Group Title

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

28. zonazoo Group Title

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

29. zepdrix Group Title

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.

30. zonazoo Group Title

well thank you again. for all your help.