# 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!

#### DrPhil

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