2. Lemniscate Curve, Find the exact coordinates of the point(s) where the tangent line to this curve is horizontal

Question

Curve:

 

(x²+y²)²=4(x²-y²)

 

Find points where tangent line is horizontal?

Answer 1

My guess is that the intent of this problem is to covert to polar form and then use the formula for finding slopes of tangent lines to an equation written in polar form.

 

It is also possible to use implicit differentiation on the equation written in rectangular form, solve for y', determine when it has a value of 0, etc. Doing so is brutal.

 

Not that the polar form is much easier.

 

Use r² = x²+y²

x=rcosθ, y=rsinθ

 

to convert to something like:

 

r⁴=4(r²)(cos² θ - sin²θ)

 

Eventually you can get this to look like:

 

r = 2 (cos2θ)^1/2

 

So r' = -2sin2θ/(cos2θ)^1/2

 

Now what?

 

The formula for dy/dθ can be derived from y=rsinθ by using the product rule. When that is equal to 0, there will be horizontal tangent lines.

 

Your book might already give you the formula for dy/dθ = r'sinθ + rcosθ.

dy/dθ = [-2sin2θ/(cos2θ)^1/2] sinθ + 2 (cos2θ)^1/2cosθ

 

Set that equal to 0, solve for theta, determine the corresponding value for r, then convert r,theta to rectangular coordinates.

 

I found these points (√6/2, √2/2) with a plus/minus combo on each one, i.e. 4 points. where tangent line is horizontal.

 

Let me know if you need any clarification.

 

Update: Jo asked how to solve for theta.

 

If you set the derivative to 0 and multiply through by (cos2Θ)^1/2 you end with something like this:

 

sin2ΘsinΘ=cos2θcosΘ

 

From here there are likely multiple ways to proceed.

 

I tried this. Divide both sides by the right side and set equal to 0.

 

This gives tan2θtanΘ - 1 = 0

 

Replace tan2Θ by its double angle equivalent. 

 

You finally end with 3tan²Θ= 1.

 

This results in Θ=pi/6 and others. Find r from r = 2(cos2Θ)^1/2, then convert to rectangular using y=rsinΘ and x = rcosΘ. (r = √2).