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!
Best Answer
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 $k^2=1$