Points on ellipse at maximal distance from center

Question

Suppose we have an equation of an ellipse: $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = c. $$

My question is where are the points situated on the ellipse at maximal distance from its center? I imagine this points are either on the major axis or on the principal axes. Is this answer correct? How should I prove it?

Edit: I would like to have a solution without Lagrange multipliers (only geometry / linear algebra techniques).

Answer 1

This can be done entirely algebraically.

Assume that $c>0$ and that $|a| < |b|$. Then $b^2 - a^2 > 0$. Now

$$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} ~=~ \dfrac{1}{b^2} (x^2+y^2) + \dfrac{b^2-a^2}{a^2b^2} x^2 ~=~ c$$

and so

$$x^2+y^2 = b^2 \left( c - \dfrac{b^2-a^2}{a^2b^2} x^2 \right)$$

Thus we must have $x^2+y^2 \le b^2c$. This upper bound is achieved precisely when $x=0$, in which case $y = \pm b \sqrt{c}$. So the points $(0,b\sqrt{c})$ and $(0,-b\sqrt{c})$ are furthest from the origin.

Likewise, if $|b| < |a|$ then the points $(a\sqrt{c}, 0)$ and $(-a\sqrt{c}, 0)$ are furthest from the origin.

So indeed, the points on the ellipse furthest to the origin are those on its major axis.

Answer 2

Geometrical view

Algebraic proof

You may rewrite your equation of ellipse as $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with a suitable substitution (replace $x$ and $y$ by $\sqrt{c}x$ and $\sqrt{c}y$ respectively). This allows you to read directly the length of the axes of the ellipse from the equation.

WLOG, assume $0 < a \le b$. (Otherwise, interchange the role of $x$ and $y$. This won't affect the objective function $d^2 = x^2 + y^2$.) Multiply both sides of the above equation by $a^2$ to get the first term of the objective function $d^2$. $$x^2 + a^2\, \frac{y^2}{b^2} = a^2$$

$$\therefore d^2 = x^2 + y^2 = a^2 - \underbrace{\left[\left(\frac{a}{b}\right)^2 - 1\right]}_{\ge 0} y^2$$

Hence, $d^2 = x^2 + y^2$ is maximized iff $y = 0$ (i.e. the point $(x,y)$ lies on either side of the longer axis of the ellipse).

Answer 3

One more:

Symmetry allows to restrict to first quadrant.

Looking for a point on the ellipse with maximum distance to origin.

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. (Rescaling your equation)

Set: $x=a\cos t$; $y=b\sin t$, $0\le t \le π/2$.

Distance square :

$d^2:= x^2+y^2=$

$a^2\cos^2 t +b^2\sin^2 t=$

$a^2\cos^2 t - b^2 \cos^2 t +b^2=$

$(a^2-b^2)\cos^2 t +b^2.$

1) $a >b$:

$d^2_{max}= (a^2-b^2)\cdot 1 +b^2=a^2$, for $t=0$.

2) $a<b$:

$d^2_{max}= (a^2-b^2)\cdot 0 +b^2= b^2$, for $t=π/2$.

(Recall $0 \le t \le π/2$ in the first quadrant).