Question

Let \(H\) be the hyperbola \(x^{2}-y^{2}=1\) and let \(S\) be the 2 -by- 2 square bisected by the asymptotes of \(H\) Let \(R\) be the anvil-shaped region bounded by the hyperbola and the horizontal lines \(y=\pm p\) (see figure). (GRAPH CANNOT COPY) a. For what value of \(p\) is the area of \(R\) equal to the area of \(S ?\) b. For what value of \(p\) is the area of \(R\) twice the area of \(S ?\)

Short answer

Question: Find the values of p such that the anvil-shaped region inside the hyperbola x^2 - y^2 = 1 and between the horizontal lines y = ±p is, respectively: (a) equal, and (b) twice the area of the square S centered at the origin, bisected by a pair of asymptotes. Answer: (a) p = 3/2; (b) p = 1.

Step-by-step solution

Find the area of square S

We know that the square lies bisected by the asymptotes of the hyperbola, \(y=\pm x\). The vertices of the square are the intersection points of asymptotes and hyperbola. To find these intersection points, we will substitute \(y\) with \(x\) and \(-x\) in the equation of the hyperbola and solve for \(x\). \(x^{2} - x^{2} = 1 \implies x=1\) and \(x^{2} - (-x)^{2} = 1 \implies x=-1\). Thus, the vertices of the square are \((1,1), (-1,1), (1,-1),\) and \((-1,-1)\). The side length of the square \(S\) is \(2\). Therefore, the area of square \(S\) is \(2^2 = 4\).

Find the area of the anvil-shaped region R

The anvil-shaped region R is a vertically symmetric region. We can find the area of the upper or lower half and then double it to get the total area. We can find the area of the upper half by integrating the hyperbola \(x^{2} - y^{2} = 1\) from \(-1\) to \(1\) and subtracting the area of the rectangle formed by \(y=p\). First, let's solve for \(x\) in terms of \(y\): \(x^{2} - y^{2} =1 \implies x =\sqrt{1+y^2}\). The area of the rectangle is \(2*(1-p)\). Now, using integration to find the upper half of R: Area (Upper-half R) \(= \int_{-1}^{1}{\sqrt{1+y^2}}dy - 2(1-p)\). Now, we can find the total area of R by multiplying the upper half area by 2: Total Area (R) \(= 2*(\int_{-1}^{1}{\sqrt{1+y^2}}dy - 2(1-p))\).

Set up equations and solve for p where area of R is equal or twice the area of S

a. To find the value of p for which the area of R is equal to the area of S, we will set up the equation and solve for p: \(2*(\int_{-1}^{1}{\sqrt{1+y^2}}dy - 2(1-p))=4\). Dividing both sides by 2 gives: \(\int_{-1}^{1}{\sqrt{1+y^2}}dy = 4+2(1-p)\). Solving for p gives: \(4-4+2(1-p) \implies p = \frac{3}{2}\). b. For the area of R to be twice the area of S, set up the equation and solve for p: \(2*(\int_{-1}^{1}{\sqrt{1+y^2}}dy - 2(1-p))=8\). Dividing both sides by 2 gives: \(\int_{-1}^{1}{\sqrt{1+y^2}}dy = 8+2(1-p)\). Solving for p gives: \(8-8+2(1-p) \implies p = 1\). In conclusion, for the area of R to be equal to the area of S, p should be \(\frac{3}{2}\), and for the area of R to be twice the area of S, p should be \(1\).

Key concepts

Hyperbolas

A hyperbola is a fascinating shape that belongs to the family of conic sections, which also includes circles, ellipses, and parabolas. Unlike the others, a hyperbola consists of two disconnected curves called branches. These curves are symmetric with respect to their axes and the center point of the hyperbola. The standard form of a hyperbola's equation, oriented along the x-axis, is \(x^2 - y^2 = c^2\), where \(c\) is a constant that determines the spread of the branches.
In the context of the hyperbola \(x^2 - y^2 = 1\), the branches open horizontally, meaning they extend along the x-axis. The asymptotes of this hyperbola, which are the lines the branches approach but never touch, are given by the equations \(y = x\) and \(y = -x\). These lines are crucial as they help in visualizing how the hyperbola behaves at large distances from its center. They also serve an essential function in determining points of intersection that aid in calculating area and symmetry.

Integration

Integration is a fundamental concept in calculus used to calculate the area under a curve. It adds up tiny slices of area beneath a graph to give the total area over a particular interval. When a function is integrated over a range, it provides the sum of infinite small quantities along an axis.
For the problem at hand, the integral \(\int_{-1}^{1}\sqrt{1+y^2}\,dy\) is used to calculate the area under the curve of the upper half of the hyperbola \(x^2 - y^2 = 1\). We integrate this function between the limits from -1 to 1, corresponding to the intersection points where the hyperbola meets its asymptotes. This integration gives us the necessary component to determine the total area under consideration. After solving the integral, we can manipulate the results to find the specific value of \(p\) for which the areas of different regions satisfy equality or other specified relationships.

Symmetry

Symmetry plays a critical role in the analysis of shapes like hyperbolas. It simplifies calculations by allowing for a focus on only one part of a shape. We can then multiply the results to find the total area or other properties for the entire shape.
The hyperbola \(x^2 - y^2 = 1\) is symmetric about the origin, evidenced by its equal behavior along diagonals defined by its asymptotes \(y=x\) and \(y=-x\). This symmetry implies that both halves above and below the x-axis are mirror images of each other. As a result, by calculating the area of one section, we automatically determine the area of the symmetric section. This property was used in the solution as it allowed us to integrate only from 0 to 1 and then multiply the result by 2 to get the entire area of the anvil-shaped region.

Area calculation

Calculating area is an integral part of understanding and solving geometrical problems. For this hyperbola-related problem, we have two main areas to consider: the square \(S\) and the anvil-shaped region \(R\).
The square \(S\) is simple; it has a side length of 2, leading to an area of \(2 \times 2 = 4\). This area forms the basis for comparing with region \(R\).
The region \(R\) requires more effort. To find its area, we first establish the limits of integration — which are the x-values from -1 to 1 — and the hyperbola equation to compute the part under the curve. We then subtract the area of the rectangle formed by \(y = p\), where the line stretches from -1 to 1 along the x-axis. This subtraction accounts for the exclusion of the rectangle's area in the overall anvil shape. By solving \(2 \times (\int_{-1}^{1}\sqrt{1+y^2}\,dy - 2(1-p))\) and setting its results equal to the area of \(S\) or twice for twice the area, we find the necessary values of \(p\) for the exercises.