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 hyperbola's equation is \( y^2 - x^2 = 1 \).

Step 1: Understand the General Form of Hyperbola

The standard form of the hyperbola centered at the origin is given by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Given the asymptotes are \( y = \pm x \), we must use the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), representing a vertical hyperbola.

Step 2: Identify Asymptotes and Parameters

The asymptotes for the hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) are \( y = \pm \frac{a}{b}x \). Given \( y = \pm x \), we have \( \frac{a}{b} = 1 \). Therefore, \( a = b \).

Step 3: Use Foci to Find Parameters

For the hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the foci are at \((0, \pm c)\) where \(c^2 = a^2 + b^2\). Given the foci \( (0, \pm \sqrt{2}) \), we get \(c = \sqrt{2}\).

Step 4: Solve for \(a^2\) and \(b^2\)

Since \( a = b \) and \( c = \sqrt{2} \), substitute into \( c^2 = a^2 + b^2 \): \[ 2 = a^2 + b^2 \] This becomes: \[ 2 = 2a^2 \] Therefore, \( a^2 = 1 \) and \( b^2 = 1 \).

Step 5: Write the Standard-Form Equation

Having \( a^2 = 1 \) and \( b^2 = 1 \), the hyperbola's equation is:\[ \frac{y^2}{1} - \frac{x^2}{1} = 1 \]This simplifies to:\[ y^2 - x^2 = 1 \].

Key Concepts

Foci of a Hyperbola

The foci of a hyperbola are two specific points that help define its shape. For a hyperbola centered at the origin with the vertical standard form equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the foci are located along the y-axis at \((0, \pm c)\). The distance \(c\) is calculated using the relationship \(c^2 = a^2 + b^2\). This means the foci are always outside the vertices since they are further from the center along the axis of the hyperbola.

In our exercise, we are given that the foci are \((0, \pm \sqrt{2})\). This indicates that \(c\) is \(\sqrt{2}\). Using the formula \(c^2 = a^2 + b^2\), this directly tells us the sum of \(a^2\) and \(b^2\) is \(2\). Determining the location of the foci is crucial for understanding the extend of the hyperbola along its axis, helping define the curve's overall structure.

Vertices of a Hyperbola

Vertices are another critical component of a hyperbola and are the closest points on the curve to its center. For our hyperbola, which has a vertical orientation, the vertices are located on the y-axis at \((0, \pm a)\). Vertices mark the points where the hyperbola intersects the line through the foci and are significant in understanding the hyperbola's width and spread.

With \(a = b\) as determined in the exercise, and given the asymptotes \(y = \pm x\) indicate that \(a = b\), we find that \(a^2 = 1\). We easily derive this from setting \(c = \sqrt{2}\) into \(c^2 = a^2 + b^2\), allowing us to solve for \(a^2\) and \(b^2\). Thus, the vertices are positioned at \((0, \pm 1)\). These points help in sketching the hyperbola and give us specific reference points on its curves.

Asymptotes of a Hyperbola

Asymptotes are imaginary lines that the branches of a hyperbola approach but never touch. They form diagonal guidelines that help to shape the opening of the hyperbola. For a hyperbola centered at the origin, the equations of the asymptotes provide critical information about its orientation and dimensions.

In the vertical form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), asymptotes are given by the equations \(y = \pm \frac{a}{b}x\). In the exercise, the asymptotes are provided as \(y = \pm x\), indicating \(\frac{a}{b} = 1\). This means that \(a = b\), simplifying our equation to have equal stretches along the axes.

These guidelines are vital for understanding the broad directional tendencies of the hyperbola's arms and help in drawing the complete figure. The asymptotes ensure that the curves of the hyperbola tend towards these lines as they extend outward, framing the space the hyperbola occupies.