Problem 53 - Thomas Calculus

Question

Exercises \(53-56\) give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes. $$ \frac{x^{2}}{4}-\frac{y^{2}}{5}=1, \quad \text { right } 2, \text { up } 2 $$

Step-by-Step Solution

Answer

New equation: \( \frac{(x-2)^2}{4} - \frac{(y-2)^2}{5} = 1 \). Center: (2, 2), Foci: (5, 2) and (-1, 2), Vertices: (4, 2) and (0, 2), Asymptotes: \( y-2 = \pm \frac{\sqrt{5}}{2}(x-2) \).

Step 1: Original Equation and Shift

We start with the original equation of the hyperbola: \( \frac{x^2}{4} - \frac{y^2}{5} = 1 \). We are told to shift it right by 2 units and up by 2 units. This involves adjusting the variables \( x \) and \( y \) by \( x - h \) and \( y - k \) respectively, where \( h = 2 \) and \( k = 2 \).

Step 2: Apply the Shift

To shift the hyperbola, substitute \( x - 2 \) for \( x \) and \( y - 2 \) for \( y \) into the original equation. The new equation becomes: \[ \frac{(x - 2)^2}{4} - \frac{(y - 2)^2}{5} = 1. \]

Step 3: New Center Calculation

The original center of the hyperbola is at \((0, 0)\). After shifting right 2 and up 2, the new center is \((2, 2)\).

Step 4: Find New Foci

The distance to each focus from the center is found using \( c = \sqrt{a^2 + b^2} \). Here, \( a^2 = 4 \) and \( b^2 = 5 \), so \( c = \sqrt{4 + 5} = 3 \). The foci are at \((2 \pm 3, 2)\), which are \((5, 2)\) and \((-1, 2)\).

Step 5: Locate New Vertices

The distance \( a \) from the center to each vertex is \( 2 \). The vertices of the hyperbola are thus at \((2 \pm 2, 2)\), which are at \((4, 2)\) and \((0, 2)\).

Step 6: Determine New Asymptotes

The equations for the asymptotes of the hyperbola are derived from the slopes \( \pm \frac{b}{a} = \pm \frac{\sqrt{5}}{2} \). Therefore, the asymptotes through the new center \((2, 2)\) are \( y - 2 = \pm \frac{\sqrt{5}}{2}(x - 2) \).

Key Concepts

Shifted Coordinate System

In mathematics, shifting a coordinate system helps to redefine the position of a shape, such as a hyperbola, on a graph. This transformation is a key step when dealing with equations of conic sections. It involves translating the original graph to a new location on the coordinate plane by adjusting both the x and y variables.
For this hyperbola, the shift involves moving it rightward by 2 units and upward by 2 units. This is accomplished by substituting the expressions \(x - 2\) for x and \(y - 2\) for y in the original equation of the hyperbola.
The revised equation becomes \[ \frac{(x - 2)^2}{4} - \frac{(y - 2)^2}{5} = 1. \] This new equation represents the hyperbola's new position in the shifted coordinate system.

Hyperbola Center

The center of a hyperbola is an important reference point that helps define its position on the coordinate plane. The center refers to the midpoint of the line segment joining the vertices of the hyperbola.
In the original hyperbola equation, the center is located at \(0, 0\). After applying the shifts detailed in the exercise, the center relocates to a new position. The transformation instructions told us to move 2 units to the right and 2 units up, so the new center becomes \(2, 2\).
Knowing the exact location of the hyperbola's center is crucial for accurately determining other elements like foci, vertices, and asymptotes.

Hyperbola Foci

The foci of a hyperbola are two special points located along the axis that help define the shape's curve. To find the foci, we calculate the distance c from the center using the relationship \(c = \sqrt{a^2 + b^2}\).
With the given hyperbola, \(a^2 = 4\) and \(b^2 = 5\), which leads to \(c = \sqrt{9} = 3\). This calculation tells us how far the foci are located on either side of the center along the x-axis, when the transverse axis is horizontal.
After shifting the hyperbola, the foci move to positions \(5, 2\) and \(-1, 2\). Understanding the location of the foci is essential since they influence the asymptotes and the overall geometry of the hyperbola.

Hyperbola Vertices

The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. These points are essential for sketching the hyperbola.
They lie on either side of the center, separated by the distance a. In this case, \(a = 2\), thus indicating that the vertices are two units away from the center along the x-axis.
After applying the 2-unit rightward and upward shifts, the vertices shift to new locations \(4, 2\) and \(0, 2\). Locating the vertices gives a better understanding of the hyperbola's orientation and scale.

Hyperbola Asymptotes

Asymptotes are lines that the hyperbola approaches but never touches. They pass through the center of the hyperbola and provide a guideline for its asymptotic behavior.
The slopes of the asymptotes are determined by \(-\frac{b}{a}\) and \(\frac{b}{a}\). With this hyperbola, the slopes are \(\pm \frac{\sqrt{5}}{2}\).
Based on the new center at \(2, 2\), the equations of the asymptotes are \[ y - 2 = \pm \frac{\sqrt{5}}{2}(x - 2). \] These equations are vital for predicting how the hyperbola will extend into the plane without intersecting the asymptotes.