Rectangle & Ellipse

Let:

• $x$ and $y$ be the sides of the rectangle
• $a$ and $b$ be half the length of the major and minor axes of the ellipse respectively
• $c$ be the distance from the center of the ellipse to its foci

The sum of the distances from the foci to a point on the ellipse is equal to the length of the major axis. Because two opposite vertices of the rectangle lie on the ellipse and the other two vertices are on the foci, we can say that:

$x\quad +\quad y\quad =\quad 2a\quad (Eq.\quad 1)$

The length of the diagonal of the rectangle is $2c$ , so $x$ , $y$ , and $2c$ form a right triangle with $2c$ being the hypotenuse.

${ x }^{ 2 }\quad +\quad { y }^{ 2 }\quad =\quad { \left( 2c \right) }^{ 2 }\quad =\quad 4{ c }^{ 2 }\quad (Eq.\quad 2)$

Squaring both sides of Eq. 1, we obtain

${ x }^{ 2 }\quad +\quad 2xy\quad +\quad { y }^{ 2 }\quad =\quad 4{ a }^{ 2 }\\ \Rightarrow \quad { x }^{ 2 }\quad +\quad { y }^{ 2 }\quad =\quad 4{ a }^{ 2 }\quad -\quad 2xy\\ \Rightarrow \quad 4{ c }^{ 2 }\quad =\quad 4{ a }^{ 2 }\quad -\quad 2xy\quad (from\quad Eq.\quad 2)\\ \Rightarrow \quad 4{ c }^{ 2 }\quad =\quad 4{ a }^{ 2 }\quad -\quad 2(200)\quad \because \quad Area\quad of\quad rectangle\quad =\quad xy\quad =\quad 200\\ \Rightarrow \quad { c }^{ 2 }\quad =\quad { a }^{ 2 }\quad -\quad 100\quad (Eq.\quad 3)$

The semi-major and minor axes and the focus line form a right triangle

${ a }^{ 2 }\quad =\quad { b }^{ 2 }\quad +\quad { c }^{ 2 }\\ \Rightarrow \quad { c }^{ 2 }\quad =\quad { a }^{ 2 }\quad -\quad { b }^{ 2 }\\ \Rightarrow \quad { a }^{ 2 }\quad -\quad { b }^{ 2 }\quad =\quad { a }^{ 2 }\quad -\quad 100\quad (from\quad Eq.\quad 3)\\ \Rightarrow \quad { b }^{ 2 }\quad =\quad 100\\ \Rightarrow \quad b\quad =\quad 10$

The area of an ellipse is given by $\pi ab$

$\pi ab\quad =\quad 200\pi \\ \Rightarrow \quad ab\quad =\quad 200\\ \Rightarrow \quad a\quad =\quad \frac { 200 }{ b } \quad =\quad \frac { 200 }{ 10 } \quad =\quad 20$

Solving for ${ c }^{ 2 }$

${ c }^{ 2 }\quad =\quad { a }^{ 2 }\quad -\quad { b }^{ 2 }\\ \Rightarrow \quad { c }^{ 2 }\quad =\quad { 20 }^{ 2 }\quad -\quad { 10 }^{ 2 }\quad =\quad 300$

Going back to Eq. 2

${ x }^{ 2 }\quad +\quad { y }^{ 2 }\quad =\quad 4{ c }^{ 2 }\\ \Rightarrow \quad { x }^{ 2 }\quad +\quad 2xy\quad +\quad { y }^{ 2 }\quad =\quad 4{ c }^{ 2 }\quad +\quad 2xy\\ \Rightarrow \quad { \left( x\quad +\quad y \right) }^{ 2 }\quad =\quad 4\left( 300 \right) \quad +\quad 2\left( 200 \right) \quad =\quad 1600\\ \Rightarrow \quad x\quad +\quad y\quad =\quad 40$

The perimeter, $P$ , of the rectangle is

$P\quad =\quad 2(x\quad +\quad y)\quad =\quad 2(40)\quad =\quad \boxed { 80 }$

x,y = sides of rectangle, so p = 2(x+y) a = length major axis b = length minor axis c = distance from center of ellipse to foci x+y = 2a ==> x^2+2xy+y^2 = 4a^2 x^2+y^2 = 4c^2 using these 2 equations we can deduce c^2=a^2-100 a^2 = b^2 + c^2 so b = 10 pi a b = area of ellipse = 200 pi ==> a = 20 c^2 = 300 ---- x^2+y^2+2xy = c^2+2(200) = 4a^2 = 4 20^2. x^2+y^2 = 4c^2 = 1200 (x+y)^2 = 4c^2+2xy = 1200+2(200) = 1600 x+y = 40 p = 2(x+y) = 2(40) = 80