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How do I find the points on a curve that are stationary?

I have the equation $$\frac{x^2}{20}+\:\frac{y^2}{5}\:=\:1$$

Using implicit differentiation I arrived at:

$$-\frac{x}{4y}$$

To find the stationary points on the curve, I made the numerator = 0 and denominator = 0, (I have no clue if this is correct or not) so I ended up with; $$-x = 0$$ and for y;
$$4y = 0$$ $$y = \frac{0}{4}\:$$ $$y = 0$$

Is the next step to take $x = 0$ and $y = 0$ and plug those values into the initial equation? I solved for y and got $\frac{\sqrt{20-x^2}}{2}$

$$\frac{0^2}{20}+\:\frac{\frac{\sqrt{(20-0^2)^2}}{2}}{5}\:=\:1$$

Checking in Desmos, an online graphing calculator, it seems the answer should be +-2.236?

Anon
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1 Answers1

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You correctly found that \begin{equation} \frac{d y}{d x} = - \frac{x}{4y} \end{equation} using implicit differentiation. The definition of a stationary point $(x^*, y^*)$ is that it lies on the curve and that \begin{equation} \left.\frac{d y}{d x}\right|_{x = x^*, y = y^*} = 0. \end{equation} Thus, we want $-\frac{x}{4y} = 0$. This happens when $x = 0$ regardless of the value of $y$. We also have to make sure that the stationary point(s) lies on the curve. To this end, plug back into the equation for the curve to see that at the stationary point(s) we must have \begin{equation} \frac{y^2}{5} = 1. \end{equation} Therefore, the two stationary points of the curve are $(0, \sqrt{5})$ and $(0, -\sqrt{5})$.

pg_star
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  • Hi pg_star, why didn't we have to solve for y, plug the x value into that, then plug 0 in to that, when x = 0? Is there a rule for stationary points that we only find x = 0 no matter what? – Anon Apr 28 '20 at 15:03
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    Valid question, let me edit my answer to make it clearer. – pg_star Apr 28 '20 at 15:07
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    “This happens when $x=0$ regardless of the value of $y$.” Not quite. If $y$ is also zero then the expression is undefined. – amd Apr 28 '20 at 19:00