Question

Rectangle ABCD has area 200. An ellipse with area $200\pi$ passes through A and C and has foci at B and D. If the perimeter of the rectangle is 16 M, then the value of M is.

• A. 5
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (A)

5

Answer 2

Let the rectangle have side lengths $l$ and $w$. Area of rectangle: $lw = 200$. Perimeter of rectangle: $2(l+w) = 16M$.

The distance between the foci B and D is $2c = \sqrt{l^2 + w^2}$. The ellipse passes through A and C. The sum of the distances from any point on the ellipse to the foci is $2a$. For point A, the distances to foci B and D are $AB = l$ and $AD = w$. Thus, $2a = l+w$.

The area of the ellipse is $\pi ab = 200\pi$, so $ab = 200$. For an ellipse, $a^2 = b^2 + c^2$. Substituting $a = \frac{l+w}{2}$ and $c = \frac{\sqrt{l^2 + w^2}}{2}$: $(\frac{l+w}{2})^2 = b^2 + (\frac{\sqrt{l^2 + w^2}}{2})^2$ $\frac{l^2 + 2lw + w^2}{4} = b^2 + \frac{l^2 + w^2}{4}$ $\frac{2lw}{4} = b^2 \implies b^2 = \frac{lw}{2}$.

Given $lw = 200$, $b^2 = \frac{200}{2} = 100$, so $b=10$. Since $ab = 200$, $a \times 10 = 200$, so $a=20$. Then $l+w = 2a = 2 \times 20 = 40$. The perimeter of the rectangle is $2(l+w) = 2(40) = 80$. We are given the perimeter is $16M$. $16M = 80 \implies M = 5$.