Problem 77 - Algebra and Trigonometry

Question

The points of intersection of the graphs of \(x y=20\) and \(x^{2}+y^{2}=41\) are joined to form a rectangle. Find the area of the rectangle.

Step-by-Step Solution

Answer

The area of the rectangle is 18 square units.

Step 1: Find the Intersection Points

First, solve the equations \(xy=20\) and \(x^{2} + y^{2} = 41\) simultaneously to get the intersection points. Since \(x = 20/y\) from the hyperbola equation, substitute that in the circle equation to obtain: \[(\frac{20}{y})^2 + y^2 = 41\]. Solving this equation gives y = ± 4 or ±5. If y = 4, x = 5 and if y = 5, x = 4. Hence the points of intersection are (5, 4), (-5, -4), (4, 5) and (-4, -5).

Step 2: Calculate the Lengths of the Rectangle's Sides

Given that the intersection points form a rectangle, use the distance formula to find the lengths of the rectangle's sides. The distance between the points (5, 4) and (4, 5) is \(\sqrt{(5-4)^2 + (4-5)^2}\) = \(\sqrt{2}\). The distance between the points (5, 4) and (-5, -4) is \(\sqrt{(5-(-5))^2 + (4-(-4))^2}\) = \(\sqrt{162}\).

Step 3: Find the Area of the Rectangle

The area of the rectangle is found by multiplying the lengths of its sides. So, the area is \(\sqrt{2} * \sqrt{162} = 18\).