Question

2(x^2 + y^2)^2 = 25(x^2 - y^2)

Find points on the lemniscate (shape of graph) where the tangent is horizontal.

Any and all help is greatly appreciated!

Answer 1

main objective is to find (x,y) when y'=0

Answer 2

I found:
\[y' = \frac{8x^3 +8xy^2-50x}{8y^3-8x^2y-50y}\]

Answer 3

...so do I set that equal to zero and solve for either x or y?

Answer 4

...so do I set that equal to zero and solve for either x or y?

Answer 5

I'm getting a little bit of sign differences in mine

Answer 6

\[y' = \frac{8x^3 +8xy^2-50x}{\color{red}{-}8y^3-8x^2y-50y}\]
is the only difference I'm seeing in my y'

Answer 7

Yep! Goddamn, my typos...

Answer 8

anyways yeah we want to find (x,y) when y'=0
and this might involved using the original equation also
we can factor out 2x on top and -2y on bottom
this will give us:
\[y'=- \frac{x}{y} \frac{4x^2+4y^2-25}{4x^2+4y^2+25} \\ \text{ so we want } x=0 \text{ or } 4x^2+4y^2-25=0\]

Answer 9

and on yeah y cannot be 0 otherwise the bottom is 0

Answer 10

so you replace the x's in the initial equation with 0 and solve for y

Answer 11

we gotta also figure out how to deal with the second equation

Answer 12

\[4x^2+4y^2=25 \\ x^2+y^2=\frac{25}{4} \]

Answer 13

we can replace x^2+y^2 in the initial equation with 25/4 ...

Answer 14

\[2(\frac{25}{4})^2=25(x^2-y^2) \\ \frac{2}{25}(\frac{25}{4})^2=x^2-y^2 \\ \frac{2(25)}{4^2}=x^2-y^2 \\ \frac{25}{8}=x^2-y^2 \]

Answer 15

I guess this gives a system of equations to solve

Answer 16

\[\frac{25}{4}=x^2+y^2 \\ \frac{25}{8}=x^2-y^2 \]

Answer 17

Here is the graph & source of the original equation.

Answer 18

you should wind up with three points

Answer 19

what does the (3,1) mean ?

Answer 20

That was the point I had to find a tangent line for, in problem 31. This is #43, and asking just for horizontal tangents, but it refers to the "lemniscate" function from #31, so I thought you might want to look at it.

Answer 21

oh okay did you see how I got the system of equations above?

Answer 22

Working on it:)

Answer 23

you might want to check my arithmetic when looking at it

Answer 24

by "replace x's in original equation", you mean 2(x^2 + y^2)^2 = 25(x^2 - y^2)

Answer 25

just for practice, you could also try changing to polar coordinates. see how simple the curve looks in polar :
\[2r^2 = 25\cos(2\theta)\]

Answer 26

yes.

Answer 27

2/25 = y^(-2) ?

Answer 28

@ganeshie8

r^2 = x^2+y^2, so the LHS should be 2(r^2)^2 which becomes 2r^4

Answer 29

@freckles your arithmetic looks good. I confirmed it with geogebra

Answer 30

\[\frac{25}{4}=x^2+y^2 \\ \frac{25}{8}=x^2-y^2 \\ \text{ gives } 2x^2=\frac{75}{8} \text{ by adding equations }\]

Answer 31

oh wait are you trying to find what y is from x=0 @amonoconnor ?

Answer 32

Ahh good catch @jim_thompson5910

Answer 33

thanks @jim_thompson5910

Answer 34

Yeah, is that wrong?

Answer 35

\[x=0 \\ 2(x^2+y^2)^2=25(x^2-y^2) \\ 2(0^2+y^2)^2=25(0^2-y^2) \\ 2y^4=-25y^2 \\ 2y^4+25y^2=0 \\ y^2(2y^2+25)=0 \\ y^2=0 \text{ and } 2y^2+25 \neq 0 \\ y=0 \\ (0,0) \text{ makes } y'=0\]

Answer 36

but

Answer 37

we said y cannot be 0

Answer 38

so (0,0) is not going to give us y'=0

Answer 39

so we will only have two points it looks like

Answer 40

or 4 lol

Answer 41

OKay... I think I follow. I get why y cannot equal zero, but I guess I don't even know why we plugged in 0 for x... and yes, there would have to be 4 with a horizontal figure eight shape, right?

Answer 42

because y'=0 when the numerator equals 0 as long as it doesn't also make the bottom 0

Answer 43

Ahh... no rise, over some run. Slope of 0. Gotcha:)

Answer 44

so you have to consider the other factor on top=0

Answer 45

\[y'=- \frac{x}{y} \frac{4x^2+4y^2-25}{4x^2+4y^2+25} \\ \text{ so we want } x=0 \text{ or } 4x^2+4y^2-25=0 \]
x=0 gives y=0 so (0,0) won't work
we look at the other factor on top
\[4x^2+4y^2-25=0 \\ 4x^2+4y^2=25 \\ x^2+y^2=\frac{25}{4}\]
plug this into the original equation if you haven't already and you should get another equation in terms of x and y giving you a system to solve

Answer 46

How does the far left chunk equal the middle?

Answer 47

far left chunk?

Answer 48

and what middle

Answer 49

Sorry, of the 3 expressions that are separated by equals signs.
4x^2 + 4y^2 − 25 = 04x^2 + 4y^2

Answer 50

These 2.

Answer 51

those aren't equal

Answer 52

are you talking these 3 equations?
\[4x^2+4y^2-25=0 \\ 4x^2+4y^2=25 \\ x^2+y^2=\frac{25}{4} \]

Answer 53

the second equation I got by adding 25 on both sides

Answer 54

Yes:) Those!

Answer 55

the third equation I was just showing I divided by 4 on both sides

Answer 56

those are three the same equation

Answer 57

I was just solving for x^2+y^2
because I see that I can easily replace x^2+y^2 in the original equation if I do so

Answer 58

do you know how got the first equation "?

Answer 59

that is the 4x^2+4y^2-25=0

Answer 60

Yes, I get that... I'm good there:) And now I can plug in 25/4 for (x^2 + y^2) ?

Answer 61

right on

Answer 62

I actually did this above
I will just copy and paste it here so you can check your work:
\[2(\frac{25}{4})^2=25(x^2-y^2) \\ \frac{2}{25}(\frac{25}{4})^2=x^2-y^2 \\ \frac{2(25)}{4^2}=x^2-y^2 \\ \frac{25}{8}=x^2-y^2\]

Answer 63

You are almost there by the way... Just a little further.

Answer 64

1/4 = (x^2 - y^2) ?

Answer 65

Then do I solve for one variable, and plug it back in?

Answer 66

\[2(x^2+y^2)^2=25(x^2-y^2) \\ \text{ we have } x^2+y^2=\frac{25}{4} \\ 2(\frac{25}{4})^2=25(x^2-y^2) \\ \text{ solving for } (x^2-y^2) \text{ by dividing both sides by 25 } \\ \frac{2}{25}(\frac{25}{4})^2=x^2-y^2\]
we try simplifying this left hand side one more time

Answer 67

try simplifying this left hand side one more time
*

Answer 68

\[\frac{2}{25}(\frac{25}{4})^2 =\frac{2}{25}\cdot \frac{25^2}{4^2} =\frac{25^2}{25} \cdot \frac{2}{4^2}=25^{2-1} \cdot \frac{2}{16}=25^{1}\cdot \frac{1}{8}=25 \cdot \frac{1}{8}=\frac{25}{8}\]

Answer 69

so we have the following two equations:
\[x^2+y^2=\frac{25}{4} \\ x^2-y^2=\frac{25}{8}\]
I would solve the system by elimination
it is already setup for eliminiantion
just add equations (eliminating y allowing you to solve for x)

Answer 70

\[2(x^2+y^2)^2 = 25(x^2-y^2)\]
\[2(\frac{25}{4})^2 = 25(x^2-y^2)\]
\[(\frac{50}{4})^2 = 25(x^2-y^2)\]
\[(\frac{1}{25})(\frac{50}{4})^2 = (x^2-y^2)\]
\[(\frac{50}{100})^2 = (x^2-y^2)\]
(1/2)^2 = (x^2 - y^2)
1/4 = (x^2 - y^2)

Answer 71

there is a square on 25 you ignored it in your third line by choosing to multiply 25 by 2

Answer 72

It wouldn't just translate in?

Answer 73

if you had 2^2 ...
you could use law of exponents and invite 2 into the square club that 25 is already in

Answer 74

that is if you had:
\[2^2(\frac{25}{4})^2=( \frac{2 \cdot 25}{4})^2 \]

Answer 75

but 2 cannot be in the square club because 2 does not have that 2 exponent thingy

Answer 76

I'm just tired, you're right lols!

Answer 77

you did the same thing with 1/25