Problem 35 - Thomas Calculus

Question

Exercises \(35-38\) give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation from the information given. $$ \begin{array}{l}{\text { Foci: }(0, \pm \sqrt{2})} \\ {\text { Asymptotes: } y=\pm x}\end{array} $$

Step-by-Step Solution

Answer

The equation of the hyperbola is \(y^2 - x^2 = 1\).

Step 1: Identify the Structure of the Hyperbola

Given that the foci are at \((0, \pm \sqrt{2})\), and the asymptotes are \(y = \pm x\), we identify that the hyperbola is vertical because the foci are on the y-axis.

Step 2: Asymptotes Information

The asymptotes are in the form \(y = \pm \frac{b}{a}x\). Since \(b = a\) (as per \(y = \pm x\)), the hyperbola's center is at the origin \((h, k) = (0,0)\), and its equation is \(\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1\).

Step 3: Use Foci to Find the Equation

The distance from the center to each focus is \(\sqrt{2}\), indicating \(c = \sqrt{2}\). For hyperbolas, \(c^2 = a^2 + b^2\). Since \(a = b\), \(a^2 + a^2 = 2a^2 = c^2\). Substitute \(c = \sqrt{2}\) to get \(2a^2 = 2\), hence \(a^2 = 1\).

Step 4: Write the Equation

We found \(a^2 = 1\), so the standard form equation of the hyperbola is \(\frac{y^2}{1} - \frac{x^2}{1} = 1\), which simplifies to \(y^2 - x^2 = 1\).

Key Concepts

Foci

The foci of a hyperbola are two distinct points that lie along its transverse axis. These points play a key role in the hyperbola's structure. In this exercise, the foci are located at (0, \pm \sqrt{2}), which indicates that the hyperbola is vertical. This is because the foci are positioned along the y-axis.To understand why we use the foci, imagine a hyperbola as it forms around these points. The definition of a hyperbola is such that the difference in distances from any point on the hyperbola to the two foci is a constant. These foci determine not just the shape of the hyperbola, but also where it opens and how wide it spreads.For vertical hyperbolas, the foci are \((0, ±c)\), where \(c\) represents the distance from the center to each focus along the y-axis. In this specific problem, \(c = \sqrt{2}\). The formula linking the foci to the hyperbola's equation is \(c^2 = a^2 + b^2\), where \(a\) and \(b\) are the distances corresponding to the terms of the standard form equation. With the foci, we can deduce much about the hyperbola's general shape and orientation.

Asymptotes

Asymptotes are invisible guiding lines that help determine the paths of the branches of a hyperbola. These lines approach the hyperbola's curves but never meet them. For the given exercise, the asymptotes are described by the equations \(y = \pm x\).
These asymptotes suggest something important about the hyperbola's symmetry and shape. They show that, at great distances from the center, the hyperbola behaves like these straight lines. The slope provides crucial values for the parameters of the hyperbola's equation.When the asymptotes are \(y = \pm \frac{b}{a}x\) and are symmetrical like \(y = \pm x\), it suggests \(b = a\). This relationship helps simplify the formula, providing key insights into the hyperbola's orientation and dimensions along both axes. Understanding the asymptotes helps predict how the hyperbola will expand and ensures symmetry is maintained throughout.

Standard-form equation

The standard form equation of a hyperbola centered at the origin takes two potential forms: horizontal or vertical. In this problem, the information provided about the foci and asymptotes leads us to a vertical hyperbola.The standard form equation of a vertical hyperbola is:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]For this particular exercise, we identify that the asymptotes lead to \(b = a\), resulting in \(\frac{y^2}{a^2} - \frac{x^2}{a^2} = 1\). With the value of \(c = \sqrt{2}\) determined through the foci, we find \(c^2 = a^2 + a^2\).

Substituting \(c = \sqrt{2}\), leads us to \(2a^2 = 2\), thus concluding that \(a^2 = 1\). Therefore, the standard form simplifies to \(y^2 - x^2 = 1\). This equation is tailored specifically to show how the hyperbola behaves in its vertical orientation, connected directly to the properties given.

Vertical hyperbolas

Vertical hyperbolas are unique because they open up and down, as opposed to left and right. This directionality stems from having foci aligned with the y-axis, guiding each branch to extend vertically.
For a vertical hyperbola, the equation is structured as:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]Features like foci and vertices aid in distinguishing a vertical hyperbola from a horizontal one. In this exercise, the foci at (0, \pm \sqrt{2}) confirm the vertical nature. Moreover, the asymptotes y = \pm x also signal its symmetry about the x-axis.
Understanding vertical hyperbolas can be particularly vital in graphing. Recognizing the symmetry along the y-axis helps in accurately sketching the curves. These hyperbolas also differ in context; while horizontal ones relate more to side-to-side stretches, vertical hyperbolas echo height-based expansion, dictated by the y-values involved. This perspective in graphing ensures a clear approach when working with related mathematical problems, facilitating visual and numeric comprehension.