# Find the points on the lemniscate where the tangent is horizontal

#### sarahjohnson

##### New member
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!

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:
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: .$$\displaystyle \left(\pm\frac{5\sqrt{3}}{8},\:\pm\frac{5}{8} \right)$$

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

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.

There's a minor typo in post#3: Change 74 to 64, below.
$$\displaystyle x^2 \;=\; \frac{75}{74}$$
[imath]\;[/imath]

Note: Post #3 was made back in 2013.

Eliz.