Find the points on the lemniscate where the tangent is horizontal

[sarahjohnson]

Find the points on the lemniscate where the tangent is horizontal.
8(x² + y²)² = 25(x² − y²)


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]

Find the points on the lemniscate where the tangent is horizontal.
8(x² + y²)² = 25(x² − y²)


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.

 

[soroban]

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]

Thank You!!! I guess my arithmetic was bad thank you very much for all of your guys' help. Saved my life

 

[Lukester101343]

Are any of you still here? Crazy how this forum has helped me almost 10 years after you had your question. Thanks for asking the question I guess. You are probably in your late 20s or 30s by now, with a family. Probably haven't done any Calculus in years. Crazy stuff.

 

[Otis]

There's a minor typo in post#3: Change 74 to 64, below.

\(\displaystyle x^2 \;=\; \frac{75}{74} \)

[imath]\;[/imath]

 

[stapel]

Note: Post #3 was made back in 2013.

Eliz.