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How do I find the point closest to origin on the eclipse:

$$x^2 + 4y^2 = 4 $$

I tried using the Lagrange multiplier method, by using $$x^2 + 4y^2 - 4 = 0$$ as a constraint, and using $$f(x,y) = x^2 + y^2$$ as the function.

Trying to use the Lagrange multiplier method I get that

$$\left\{\begin{array}{l}2x = λ2x \\ 2y = λ8y\\ x^2 +4y^2 = 4\end{array}\right.$$

Trying to solve the equations for λ, I get that λ is both 1 and 1/4.

Thank you very much!

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You set up the equations correctly! To solve them, note that if $\lambda=1$ we get
$$2x=2x$$ $$2y=8y$$ Therefore $y=0$ (and $x$ must be $\pm 2$). And if $\lambda=\frac14$ we get

$$2x=\frac12x$$ $$2y=2y$$ Therefore $x=0$ (and $y$ must be $\pm 1$).

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  • $\begingroup$ I see. Could you explain to me how you get x = +-2 from 2x = 2x? $\endgroup$ – Sonofgreek Apr 6 '19 at 10:51
  • $\begingroup$ @dondeman: that comes from substituting $y=0$ in the equation $x^2 + 4y^2 - 4 = 0$. $\endgroup$ – TonyK Apr 6 '19 at 10:52
  • $\begingroup$ Sorry for my stupidity, how do you get y = 0, if I put λ = 1, I get 2x = 2x, and 2y = 8y, solving these I get x = x, and y = 4y , which makes no sense to me. $\endgroup$ – Sonofgreek Apr 6 '19 at 10:55
  • $\begingroup$ $y=4y$ has the unique solution $y=0$... $\endgroup$ – TonyK Apr 6 '19 at 10:56
  • $\begingroup$ I see that now. Sorry I was just being really stupid. Thank you very much for your help $\endgroup$ – Sonofgreek Apr 6 '19 at 11:00
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If the Lagrange multiplier method is not mandatory,

Method$\#1:$

Any point on the ellipse $P(2\cos t,\sin t)$

If the distance of $P$ from the origin is $d\ge0$

$$d^2=4\cos^2t+\sin^2t=3\cos^2t+1$$

Now $0\le\cos^2t\le1$ as $t$ is real

Method$\#2:$

If $(h,k)$ be any point, $h^2+4k^2=4\iff h^2=?, 4k^2=4-h^2\le4\iff k^2\le1$

We need to minimize $h^2+k^2=4-3k^2\ge4-3$

which occurs if $k^2=1$

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No, you don't get that $\lambda$ is $1$ and $\frac14$. What you get is that $\lambda$ is $1$ or $\frac14$. The points of the ellipsis that you get are $\left(0,\pm1\right)$ and $(\pm2,0)$. Of these four points, the ones that are closest to the origin are $\left(0,\pm1\right)$, of course.

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  • $\begingroup$ You mean $(0,\pm 1)$ and $(\pm 2,0)$. $\endgroup$ – TonyK Apr 6 '19 at 10:51
  • $\begingroup$ I've edited my answer. Thank you. $\endgroup$ – José Carlos Santos Apr 6 '19 at 10:54
  • $\begingroup$ Sorry for my stupidity how do you get those numbers? Solving with λ = 1, I get that 2x = 2x and 2y = 8y. With λ = 1/4 I get that 2x = 1/2x, and 2y = 2y. Edit: I get that now, thank you very much for your help $\endgroup$ – Sonofgreek Apr 6 '19 at 10:57
  • $\begingroup$ It seems that you forgot the third equation. Suppose that $\lambda=1$. Then you get the trivial equality $2x=2x$. You also get $2y=8y$, from which you deduce that $y=0$. And then you deduce from $x^2+4y^2=4$ (and from $y=0$) that $x=\pm2$. $\endgroup$ – José Carlos Santos Apr 6 '19 at 11:01
  • $\begingroup$ I did. Thank you very much to both you and Tony K $\endgroup$ – Sonofgreek Apr 6 '19 at 11:03

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