# Find point closest to origin on ellipse

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!

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 $$\pm 1$$).

• I see. Could you explain to me how you get x = +-2 from 2x = 2x? – Sonofgreek Apr 6 '19 at 10:51
• @dondeman: that comes from substituting $y=0$ in the equation $x^2 + 4y^2 - 4 = 0$. – TonyK Apr 6 '19 at 10:52
• 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. – Sonofgreek Apr 6 '19 at 10:55
• $y=4y$ has the unique solution $y=0$... – TonyK Apr 6 '19 at 10:56
• I see that now. Sorry I was just being really stupid. Thank you very much for your help – Sonofgreek Apr 6 '19 at 11:00

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 $$k^2=1$$

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.

• You mean $(0,\pm 1)$ and $(\pm 2,0)$. – TonyK Apr 6 '19 at 10:51
• I've edited my answer. Thank you. – José Carlos Santos Apr 6 '19 at 10:54
• 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 – Sonofgreek Apr 6 '19 at 10:57
• 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 $x=\pm2$. – José Carlos Santos Apr 6 '19 at 11:01
• I did. Thank you very much to both you and Tony K – Sonofgreek Apr 6 '19 at 11:03