Question

The curve with the equation 2(x^2+y^2)^2=25(x^2-y^2) is called a lemniscate
a) Find y'
b) Find the equation of the tangent line to the lemniscate at the point (-3,1)
c) Find the points on the lemniscate where the tangent line is horizontal.

Answer 1

I got a) and b), but I don't know what to do for c). Please help!
a) \[\frac{ dy }{ dx }=\frac{ x(25-4(x^2+y^2)) }{ y(25+4(x^2+y^2)) }\]
b) 9x-13y+40=0

Answer 2

\(\large y'(x)=\LARGE\frac {-x(4x^2+4y^2-25)}{y(4x^2+4y^2+25)}\)

to find all the points where the tangent is horizontal, you set the derivative numerator equal to zero but the denominator doesn't equal to zero

to find all the points where the tangent is vertical, you set the derivative denominator equal to zero but the numerator doesn't equal to zero.

Answer 3

your question is asking for horizontal tangent. So solve that and substitute back to the original equation

Answer 4

solve for y, but since your original equation has \(2(x^2+y^2)^2=25(x^2-y^2)\)
since the y's are y^2, make the equation equal to y^2 so you can easily substitute it back in

Answer 5

I got this y^2=25/4+x^x? So I sub this equation to the original equation? @Zale101

Answer 6

Yes

Answer 7

Opps i meant y^2=-25/4+x^2

Answer 8

and solve for x, after your solved for x, you'll get some number
plug the number in x of this equation y^2=-25/4+x^2 and solve for y

and from there, you'll have x and y (points)

Answer 9

Is there an easier way to solve for this after subbing the equation? I got this for now:
\[2(x^2+\frac{ 25 }{ 4 }+x^2)^2=25(x^2+\frac{ 25 }{ 4 }+x^2)\]

Answer 10

Well, i dont know of any easier method, but what you did looks alright to me

Answer 11

I meant 2(x^2 - 25/4+x^2)^2...

Answer 12

I got x=5/4?

Answer 13

I think my answer is wrong :(

Answer 14

http://www.wolframalpha.com/input/?i=2%28x2%2B25%2F4%2Bx2%292%3D25%28x2-25%2F4%2Bx2%29

Answer 15

wolfram has something else :T

Answer 16

My teacher told us that \[x=\pm \frac{ 5\sqrt{3} }{ 4 }\]

Answer 17

can any of you guys help me with my math

Answer 18

hummm...

Answer 19

I can't get the answer

Answer 20

thisis what i keep getting
\(−x(4x^2+4y^2−25)=0\)
\(y^2=25-x^2 \)

\(2(x^2+(\Large \frac{25}{4}-x^2))=25(x^2-(\Large\frac{25}{4}-x^2))\)

\(x=\Large\frac{3\sqrt{\frac{3}{2}}}{2}\)

Answer 21

Where did you get the negative sign from -x(4x^2+4y^2-25)=0?

Answer 22

\(2\frac{d}{dx}((x^2+y^2)^2=25\frac{d}{dx}(x^2-y^2)\)

\((4x^2+4y^2)*(2x+2y*y')=25(2x-2y*y')\)
simplify
\(8x^3+8x^2y*y'+8xy^2+8y^3*y'=50x-50y*y'\)

\(8x^2y*y'+8y^3*y'+50y*y=50x-8xy^2-8x^3\)

\(8x^2y*y'+8y^3*y'+50y*y=-8x^3-8y^2x+50x\)

\(y'=\Large\frac{-8x^3-8y^2x+50x}{8y^3+8x^2y+50y}\)

Answer 23

factor out a common factor, which is 2
\(y'=\Large\frac{-2x(4y^2+4x^2-25)}{2y(4y^2+4x^2+25)}=\Large\frac{-x(4y^2+4x^2-25)}{y(4y^2+4x^2+25)}\)

Answer 24

Hmmm... The back of the book, there wasn't negative sign...

Answer 25

(shrugs) :/

Answer 26

i'm sure the implicit differentiation i did is correct. There must be something wrong....

Answer 27

okay, thanks for helping me out! :) @Zale101