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How do you find the number of critical points of an implicit equation such as $xy(x-6y)=9a^3$ ?

I have managed to differentiate and get $$\frac{dy}{dx} = 6y^2 - 2xy.$$ I don't know if I'm on the right route.

amWhy
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Jaimie1000
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  • critical points – Jaimie1000 Jul 04 '18 at 15:33
  • Are you sure that derivative is correct? Looks to me like you’ve only gotten its numerator. – amd Jul 04 '18 at 18:29
  • When you get the first derivative and equate it t zero, then cross multiply to get an equation in terms of x and y, that is what you get. I do realize that I did not put the denominator and equate to zero, sorry my bad and thanks for the correction. – Jaimie1000 Jul 05 '18 at 12:52

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You can use the implicit derivative: $$2xy+x^2y'-6y^2-6x\cdot 2yy'=0$$

assuming that $$y=y(x)$$ or you get the explicit function:

$$y_{1,2}=\frac{1}{12}x\pm\sqrt{\frac{1}{144}x^2-\frac{3}{2}a^3}$$ and compute the derivative directly

Dr. Sonnhard Graubner
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  • Thank you so much for the first suggestion but it will lead to the point i managed to get to. a little more explanation after that point will be helpful. – Jaimie1000 Jul 04 '18 at 17:35
  • Using the implicit derivative did help me thank you. I got y=0 and x=3y. when replaced in the equation, only x=3y works to give a solution. – Jaimie1000 Jul 05 '18 at 12:56