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

2(x^2 + y^2)^2 = 25(x^2 − y^2)

Question

Find the points on the lemniscate where the tangent is horizontal. 

2(x^2 + y^2)^2 = 25(x^2 − y^2)

anonymous · Answer

@zepdrix

anonymous · Answer

Familiar with implicit differentiation?

anonymous · Answer

I know I have to use it I just struggle with it.

anonymous · Answer

ok, well the plan is to use implicit differentiation to solve for dy/dx, then solve for (x,y) : dy/dx = 0.

anonymous · Answer

Remember that implicit differentiation means using the chain rule. What is being implied is that y =f(x), so any derivative of y must include y'

anonymous · Answer

i need 4 points....

zepdrix · Answer

Did you find your derivative yet toots? :o

anonymous · Answer

yes sir!

anonymous · Answer

$$4(x^2+y^2)(2x+2y*y')=25(x^2-y^2)(2x-2y*y')$$

zepdrix · Answer

The initial expression, was the right side supposed to be squared as well?

anonymous · Answer

no lol

zepdrix · Answer

Ok then we wouldn't have this,$$4(x^2+y^2)(2x+2y*y')=25\cancel{(x^2-y^2)}(2x-2y*y')$$

zepdrix · Answer

You took the derivative of that, that's what the other brackets are.

anonymous · Answer

$$25(2x-2y*y')$$

zepdrix · Answer

Hmmm I'm trying to decide whether we should solve for y' or just plug in y'=0 from here.

anonymous · Answer

you are the boss! :)

zepdrix · Answer

Ya let's just plug in y'=0 from here.
I'm thinking it will work out nicely :o

zepdrix · Answer

y' represents the slope of the lines tangent to the curve.
When y'=0 the slope of the tangent is zero (horizontal).

zepdrix · Answer

So we need to find the locations at which that is happening.

anonymous · Answer

$$4(x^2+y^2)(2x)=25(2x)$$

anonymous · Answer

i did not even see you put that equation up lol

zepdrix · Answer

XD my bad, stealing all the fun. I'll erase it hehe

anonymous · Answer

you didn't have to lol i was happy i got the same thing

anonymous · Answer

now do i solve for y?

usukidoll · Answer

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