Find point closest to origin on ellipse

Question

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!

Answer 1

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 $\pm 1$).

Answer 2

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 $k^2=1$

Answer 3

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.