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Chapter 8: Problem 77

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.

Short Answer

Expert verified
The area of the rectangle is 18 square units.

Step by step solution

01

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 = 卤 4 or 卤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).
02

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 \(\sqrt{(5-(-5))^2 + (4-(-4))^2}\) = \(\sqrt{162}\).
03

Find the Area of the Rectangle

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

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