Problem 81 - Precalculus: Functions and Graphs

Question

Graph the hyperbolas on the same coordinate plane, and determine the number of points of intersection. $$\frac{(x-0.3)^{2}}{1.3}-\frac{y^{2}}{2.7}=1 ; \frac{y^{2}}{2.8}-\frac{(x-0.2)^{2}}{1.2}=1$$

Step-by-Step Solution

Answer

The hyperbolas intersect at two points.

Step 1: Identify the Equations of the Hyperbolas

The given hyperbolas are 1. \(\frac{(x-0.3)^{2}}{1.3} - \frac{y^{2}}{2.7} = 1\), which is in the standard form of a hyperbola centered at \((0.3, 0)\), opening horizontally.2. \(\frac{y^{2}}{2.8} - \frac{(x-0.2)^{2}}{1.2} = 1\), which is in the standard form of a hyperbola centered at \((0.2, 0)\), opening vertically.

Step 2: Plot the First Hyperbola

To graph the first hyperbola \(\frac{(x-0.3)^{2}}{1.3} - \frac{y^{2}}{2.7} = 1\), identify the center \((0.3, 0)\), the distance \(a=\sqrt{1.3}\) along the x-axis, and the distance \(b=\sqrt{2.7}\) along the y-axis.Draw the box outlined by these distances, and sketch the hyperbola opening horizontally from this center.

Step 3: Plot the Second Hyperbola

For the second hyperbola \(\frac{y^{2}}{2.8} - \frac{(x-0.2)^{2}}{1.2} = 1\), identify the center \((0.2, 0)\), the distance \(a=\sqrt{2.8}\) along the y-axis, and the distance \(b=\sqrt{1.2}\) along the x-axis.Draw the box outlined by these distances, and sketch the hyperbola opening vertically from this center.

Step 4: Determine Points of Intersection

Examine the plotted graphs to determine where they intersect. The equations have been designed to have horizontal and vertical intersections depending on the symmetry of each hyperbola. The intersections will occur where the plots of the hyperbolas cross each other.

Key Concepts

Equation of Hyperbola

A hyperbola is a set of all points such that the difference of the distances from two fixed points called foci is constant. The standard equation of a hyperbola can either focus horizontally or vertically.
For a horizontally oriented hyperbola, the equation takes the form:

• \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)

While for a vertically oriented hyperbola, the equation is:

• \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Here, \((h, k)\) is the center of the hyperbola where it is symmetric, "a" represents the distance from the center to each vertex along the transverse axis, and "b" is the distance from the center to each endpoint of the conjugate axis. Understanding these components is crucial because they define the overall shape and opening direction of the hyperbola.
In the given exercise, the first hyperbola's equation is \(\frac{(x-0.3)^2}{1.3} - \frac{y^2}{2.7} = 1\), indicating a horizontal hyperbola, and the second equation \(\frac{y^2}{2.8} - \frac{(x-0.2)^2}{1.2} = 1\) represents a vertical hyperbola. Each differs in orientation but shares similar mathematical principles.

Intersection Points

Intersection points are locations where the graphs of the hyperbolas crossing each other on the same coordinate plane. Determining these points is important to solving systems of equations involving hyperbolas.
By plotting the graphs of hyperbolas, we can visually identify their intersection points. However, finding exact intersection coordinates might require solving the equations simultaneously.
When graphed, hyperbolas may intersect at points where coordinates satisfy both hyperbola equations.

• Graph the hyperbolas based on their given equations.
• Identify potential intersection points visually.
• Verify those points satisfy both equations.

In our exercise, each plotted hyperbola will intersect due to symmetric properties and designed distances, showing points of intersection where both curves meet.

Horizontal Hyperbola

A horizontal hyperbola opens to the left and right along the x-axis. It takes the form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). The center\((h, k)\) defines where the two branches of the hyperbola are centered. Horizontal hyperbolas have:

• Vertices at \((h \, \pm \ a, k)\)
• Foci farther along the x-axis

The separation of branches is primarily influenced by their distance "a", while "b" helps dictate the hyperbola's "spread" across the y-direction.
In our exercise, the equation \(\frac{(x-0.3)^2}{1.3} - \frac{y^2}{2.7} = 1\) describes a horizontal hyperbola with a center at \((0.3, 0)\). Taking the square root of the denominator, "a" is approximately \(\sqrt{1.3}\), setting the distance horizontally from the center to each vertex, defining how wide it opens.

Vertical Hyperbola

A vertical hyperbola opens upwards and downwards along the y-axis. This form is given as \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \). Its center \((h, k)\) indicates the midpoint between its vertices along the y-axis. Vertical hyperbolas have:

• Vertices at \((h, k \, \pm \ a)\)
• Foci farther along the y-axis

The primary stretch is given by the "a" value, influencing the vertical spread.
For the exercise at hand, the equation \(\frac{y^2}{2.8} - \frac{(x-0.2)^2}{1.2} = 1\) is a vertical hyperbola centered at \((0.2, 0)\). Here, "a" relates to the vertical distance of approximately \(\sqrt{2.8}\), determining how much it opens outward above and below that center point.