1. ## Ellipse Probability.

Question:

A rhombus is drawn connecting the endpoints of the major and minor axes of the ellipse $\displaystyle (\frac{x}{911})^2+(\frac{y}{2000})^2=1$. A point is chosen at random from inside the ellipse. What is the probability that the point is also inside the rhombus.

Well, I'll go ahead and state a few formulas, before I begin my work:

(pi)(a)(b) = the area of an ellipse.
(the product of the diagonals)/2 = the area of a rhombus.

So, we have a vertical ellipse, with a = 2000, and b = 911. The diagonals of the rhombus are 2(2000) and 2(911). Therefore, we have the area of the rhombus to be 2(2000)(911), using the formula from above. The area of the ellipse will be (pi)(2000)(911). Therefore, we have the probability as $\displaystyle \frac{(2)(2000)(911)}{(\pi)(2000)(911)} = \frac{2}{\pi}$

Would this solution and answer be correct?

2. Originally Posted by rtblue
Question:

A rhombus is drawn connecting the endpoints of the major and minor axes of the ellipse $\displaystyle (\frac{x}{911})^2+(\frac{y}{2000})^2=1$. A point is chosen at random from inside the ellipse. What is the probability that the point is also inside the rhombus.

Well, I'll go ahead and state a few formulas, before I begin my work:

(pi)(a)(b) = the area of an ellipse.
(the product of the diagonals)/2 = the area of a rhombus.

So, we have a vertical ellipse, with a = 2000, and b = 911. The diagonals of the rhombus are 2(2000) and 2(911). Therefore, we have the area of the rhombus to be 2(2000)(911), using the formula from above. The area of the ellipse will be (pi)(2000)(911). Therefore, we have the probability as $\displaystyle \frac{(2)(2000)(911)}{(\pi)(2000)(911)} = \frac{2}{\pi}$

Would this solution and answer be correct?
Yes.