Find the points on the lemniscate where the tangent is horizontal

sarahjohnson

New member
Joined
Jul 20, 2013
Messages
20
Find the points on the lemniscate where the tangent is horizontal.
8(x2 + y2)2 = 25(x2y2)


I got the derivative is
y'(x) = (x (-16 x^2-16 y^2+25))/(y (16 x^2+16 y^2+25))

and then I tried to set top equal to 0 and solve for x^2+y^2 which =25/16

then I plugged this back into the original equation to find x=+or- 15/sqrt(32)

the problem I have is finding y

When I try to plug x back into x^2-y^2 I get a negative root which is undefined?

I get the overall method, just can't get the answer.

Help would be much appreciated. Thanks!

 

DrPhil

Senior Member
Joined
Nov 29, 2012
Messages
1,383
Find the points on the lemniscate where the tangent is horizontal.
8(x2 + y2)2 = 25(x2y2)


I got the derivative is
y'(x) = (x (-16 x^2-16 y^2+25))/(y (16 x^2+16 y^2+25))

and then I tried to set top equal to 0 and solve for x^2+y^2 which =25/16
...OK this far

then I plugged this back into the original equation to find x=+or- 15/sqrt(32)

the problem I have is finding y

When I try to plug x back into x^2-y^2 I get a negative root which is undefined?

I get the overall method, just can't get the answer.

Help would be much appreciated. Thanks!

Since x and y appear only as squares, I made substitutions \(\displaystyle u = x^2, v = y^2\). That made the derivatives much easier to write!

\(\displaystyle \dfrac{dv}{du} = \dfrac{25 - 16(u + v)}{25 + 16(u+v)} = 0 \implies u+v = \dfrac{25}{16}\)

Plugging that into the original equation gives an equation for \(\displaystyle (u - v)\), so you have two equations in two unknowns:

\(\displaystyle u+v = \dfrac{25}{16}\)

\(\displaystyle u-v = \dfrac{25}{32}\)

Solve that system to find points that are on the lemniscate AND have horizontal tangent.
 
Last edited:

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,587
Hello, sarahjohnson!

Find the points on the lemniscate where the tangent is horizontal.
. . \(\displaystyle 8(x^2+y^2)^2 \:=\:25(x^2-y^2)\;\;[1]\)

I got the derivative is: .\(\displaystyle y' \:=\:\dfrac{x[25-16(x^2+y^2)]}{y[25 + 16(x^2+y^2)]} \) . I agree!

I set the numerator equal to 0 and got: .\(\displaystyle x^2+y^2 \:=\:\frac{25}{16}\) . Me too!

then I plugged this back into the original equation to find: .\(\displaystyle x\:=\:\pm \frac{15}{\sqrt{32}}\) . no . . .

We have: .\(\displaystyle x^2+y^2\:=\:\frac{25}{16}\;\;[2]\)

Substitute into [1]: .\(\displaystyle 8\left(\frac{25}{16}\right)^2 \:=\:25(x^2-y^2)\)

. . \(\displaystyle \dfrac{\color{red}{\rlap{/}}{8}\cdot25\cdot\color{green}{\rlap{//}}25}{16\cdot\color{red}{\rlap{//}}16_2} \:=\:\color{green}{\rlap{//}}25(x^2-y^2) \quad\Rightarrow\quad x^2-y^2 \:=\:\frac{25}{32} \;\;[3]\)

Add [2] and [3]: .\(\displaystyle 2x^2 \:=\:\frac{75}{32} \quad\Rightarrow\quad x^2 \:=\:\frac{75}{74} \)

. . Hence: .\(\displaystyle x \:=\:\pm\frac{5\sqrt{3}}{8}\)

Substitute into [2]: .\(\displaystyle \frac{75}{64} + y^2 \:=\:\frac{50}{32} \quad\Rightarrow\quad y^2 \:=\:\frac{25}{64}\)

. . Hence: .\(\displaystyle y \:=\:\pm\frac{5}{8}\)


There are four horizontal tangents.
. . They occur at: .\(\displaystyle \left(\pm\frac{5\sqrt{3}}{8},\:\pm\frac{5}{8} \right)\)
 

sarahjohnson

New member
Joined
Jul 20, 2013
Messages
20
Thank You!!! I guess my arithmetic was bad :p thank you very much for all of your guys' help. Saved my life
 
Top