Problem 30 - Thomas Calculus

Question

Exercises \(27-34\) give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. $$ y^{2}-x^{2}=4 $$

Step-by-Step Solution

Answer

The asymptotes are \(y = \pm x\) and foci are at \((0, \pm 2\sqrt{2})\).

Step 1: Identify the Given Equation

The given equation is \( y^2 - x^2 = 4 \). This is a hyperbola since it is a difference of squares.

Step 2: Rearrange to Standard Form

The standard form for a hyperbola is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). We need to divide through by 4: \[ \frac{y^2}{4} - \frac{x^2}{4} = 1 \].

Step 3: Identify Constants and Calculate Asymptotes

In the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), we identify \( a^2 = 4 \) and \( b^2 = 4 \). Thus, \( a = 2 \) and \( b = 2 \). The asymptotes for a hyperbola are given by \( y = \pm \frac{a}{b}x \). Here, \( y = \pm x \) are the asymptotes.

Step 4: Determine the Foci

The foci for a hyperbola are located at a distance \( c \) from the center, where \( c^2 = a^2 + b^2 \). Calculate \( c^2 = 4 + 4 = 8 \), so \( c = \sqrt{8} = 2\sqrt{2} \). The foci are located on the y-axis at \((0, \pm 2\sqrt{2})\).

Step 5: Sketch the Hyperbola

Draw the coordinate axes and plot the center at the origin (0,0). Draw the rectangular box determined by \( a = 2 \) and \( b = 2 \) along the axes. Sketch the asymptotes \( y = x \) and \( y = -x \). Mark the foci at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\). Draw the hyperbolic branches opening upwards and downwards intersecting the box sides vertically.

Key Concepts

Standard Form of Hyperbola

The standard form is crucial for identifying and analyzing the characteristics of a hyperbola. For hyperbolas with vertical transverse axes, like the one in our exercise, the standard form is given by:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]This equation highlights the relationship between the variables and shows the orientation of the hyperbola. To transform any given hyperbolic equation into this form, you must manipulate it so it equals 1 on the right-hand side. In our given problem, we had:\[ y^2 - x^2 = 4 \]To rearrange this into standard form, divide every term by 4 to achieve:\[ \frac{y^2}{4} - \frac{x^2}{4} = 1 \]Doing this identifies that both the horizontal and vertical distances (\(a\) and \(b\)) from the center point to the vertices are equal to 2, since \(a^2 = b^2 = 4\). Understanding this form helps in drawing and analyzing hyperbolas.

Asymptotes of Hyperbola

Asymptotes are invisible guide lines that hyperbolas approach as they extend into infinity. They help in determining the slope and orientation of hyperbolic branches. For hyperbolas represented by:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]The equations of the asymptotes are derived from the slope \( \pm \frac{a}{b} \). In our case:\[ a = b = 2 \]Thus, the asymptotes are:\[ y = \pm x \]These lines pass through the origin and are symmetrical about the \(x\) and \(y\) axes. The presence of asymptotes aids in drafting a more accurate hyperbola, ensuring that the curves correctly approach these lines as they move outward from the center.

Foci of Hyperbola

In a hyperbola, the foci are two specific fixed points and play a crucial role in its definition and shape. They lie along the transverse axis. The distance from the center to each focus can be calculated using:\[ c^2 = a^2 + b^2 \]Given \(a^2 = 4\) and \(b^2 = 4\) in our problem, we find:\[ c^2 = 4 + 4 = 8 \]\[ c = \sqrt{8} = 2\sqrt{2} \]So, the coordinates of the foci for this hyperbola are \((0, \pm 2\sqrt{2})\). These foci are aligned along the y-axis because the hyperbola opens vertically. They are essential in defining the shape of a hyperbola and understanding its properties. In sketches, these points are marked and play a key role in shaping the hyperbola around them.