# Thread: Derifferentiating a "lemniscate"

1. ## Derifferentiating a "lemniscate"

"Find the points on the lemniscate $(x^2+y^2)=25(x^2-y^2)$"

I found the derivative of the equation, and solved for $y'$, getting $(x^2+y^2)=\frac {25} {4}$, but i don't know where to go from here. I was given a couple hints, one of which being "Is the observation $x^2-y^2=(x^2+y^2)-2y^2$ helpful? And I don't think it is, but it's obvious where they got it--from $x^2+y^2=x^2+y^2$

2. Originally Posted by Chokfull
"Find the points on the lemniscate $(x^2+y^2)=25(x^2-y^2)$" Mr F says: Points that satisfy what condition?

I found the derivative of the equation, and solved for $y'$, getting $(x^2+y^2)=\frac {25} {4}$, but i don't know where to go from here. I was given a couple hints, one of which being "Is the observation $x^2-y^2=(x^2+y^2)-2y^2$ helpful? And I don't think it is, but it's obvious where they got it--from $x^2+y^2=x^2+y^2$
$(x^2+y^2)=25(x^2-y^2)$ .... (1)

dy/dx = whatever .... (2)

Solve equations (1) and (2) simultaneously.

3. Right, sorry, i never finished my sentence :P
I meant the points at which the tangent line has a slope of 0

4. Hello Chokfull

I think you also missed out a power of 2. The equation should be
$(x^2+y^2)^2 = 25(x^2-y^2)$
So, when we differentiate:
$2(x^2+y^2)\left(2x+2y\frac{dy}{dx}\right)=50x -50y\frac{dy}{dx}$
and putting $\frac{dy}{dx}=0$ gives:
$x^2+y^2 = \frac{25}{2}$ (I think - not $\frac{25}{4}$)
Now we put this back into the original equation, using the hint that $x^2-y^2 = (x^2+y^2)-2y^2$:
$\left(\frac{25}{2}\right)^2=25\left(\frac{25}{2}-2y^2\right)$
which you can now solve for $y$, and hence for $(x,y)$.

5. I see, thanks a lot! Now i didn't see before that you could replace the $x^2-y^2$ with $x^2+y^2$ and use that to get a contant in place of every x--thus allowing me to solve for y!