Question

Rectangle $ABCD$ has area $200$. An ellipse with area $200\pi$ passes through $A$ and $C$ and has foci at $B$ and $D$. Find the perimeter of the rectangle.

Answer 1

Step-1: Forming the equations:

Given the area of the rectangle is $200$

Let the length be $x$ and the breadth be $y$

We know that the length of the major axis of the ellipse is $2a$ and the length of the minor axis is $2b$

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

Given the area of the ellipse is $200\pi$

$$\begin{gathered} \Rightarrow \pi ab=200\pi \\ \Rightarrow ab=200 \end{gathered}$$

From the area of the rectangle we have

$\Rightarrow xy=200$

From geometry, we know that the length of the diagonal of the rectangle can be given as $\sqrt{x^2+y^2}$ and $x+y=2a$

Step- 2: Comparing the length of the axes and the distance from the foci to the center:

$\begin{array}{rcl}{\left(x+y\right)}^2& =& 4a^2\\ x^2+y^2& =& 4a^2{e}^2=4a^2-4b^2\left[{\text{∵e is eccentricity of ellipse,a}}^2{\text{e}}^2{\text{=a}}^2{\text{-b}}^2\right]\\ & \Rightarrow & x^2+y^2+2xy=x^2+y^2+4b^2\\ & \Rightarrow & 2xy=4b^2\\ & \Rightarrow & 2·200=4b^2\left[\because xy=200\right]\\ & \Rightarrow & b=\sqrt{100}\\ & \Rightarrow & b=10\\ \therefore ab& =& 200\\ & \Rightarrow & a=20\end{array}$

We have $x+y=2a$

$$\begin{gathered} \Rightarrow x+y=40 \\ \Rightarrow x+\frac{200}{x}=40\left[\because xy=200\right] \\ \Rightarrow x^2-40x+200=0 \end{gathered}$$

Therefore the roots of the quadratic equation is the side of the rectangle

We know that for a quadratic equation $a{x}^2+bx+c=0$, the sum of the root is given by $-\frac{b}{a}$

Step-3: Comparing the standard form of the quadratic equation with having the sum of the root is $40$

So the sum of the side of the rectangle is $40$

Therefore the perimeter of rectangle is $2\times$$($Sum of the length and the breadth$)$

Therefore the perimeter of the rectangle is $2\times 40=80unit$

Hence, the perimeter of the rectangle is $80$unit.