Finding The Equation And Graph Of A Hyperbola Given Six Points

In this article, we will determine the equation of a hyperbola given specific points it passes through and illustrate its graph. The provided points are $(2 \pm 3 \sqrt{2}, \pm 2)$, $(5,0)$, and $(-1,0)$. This exercise combines algebraic techniques with geometric understanding to solve a classic problem in analytic geometry.

Understanding Hyperbolas

Before diving into the solution, let's briefly review the basics of hyperbolas.

A hyperbola is a type of conic section formed by the intersection of a plane and a double cone. It consists of two symmetrical branches that open away from each other. The key features of a hyperbola include:

• Center: The midpoint between the two foci.
• Foci: Two fixed points inside the hyperbola that define its shape.
• Vertices: The points where the hyperbola intersects its transverse axis.
• Transverse Axis: The axis that passes through the foci and vertices.
• Conjugate Axis: The axis perpendicular to the transverse axis, passing through the center.
• Asymptotes: Lines that the hyperbola approaches as it extends to infinity.

The standard equation of a hyperbola centered at $(h, k)$ depends on whether the transverse axis is horizontal or vertical:

• Horizontal Transverse Axis: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$
• Vertical Transverse Axis: $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$

Here, $a$ is the distance from the center to each vertex, and $b$ is related to the distance from the center to the co-vertices (endpoints of the conjugate axis). The relationship between $a$, $b$, and the distance $c$ from the center to each focus is given by $c^2 = a^2 + b^2$.

Determining the Hyperbola's Equation

Initial Assessment

The given points are $(2 + 3 \sqrt{2}, 2)$, $(2 + 3 \sqrt{2}, -2)$, $(2 - 3 \sqrt{2}, 2)$, $(2 - 3 \sqrt{2}, -2)$, $(5, 0)$, and $(-1, 0)$.

Notice that the points $(5, 0)$ and $(-1, 0)$ lie on the x-axis. These points are particularly valuable because they give us immediate clues about the hyperbola's orientation and parameters.

The fact that the hyperbola passes through both $(5,0)$ and $(-1,0)$ indicates that the transverse axis is horizontal (since these points are horizontally aligned). These points are also the x-intercepts of the hyperbola, which are crucial for determining the center and the value of $a$.

Finding the Center and Vertices

The center of the hyperbola is the midpoint of the segment connecting the vertices (or the x-intercepts in this case). Let's calculate the midpoint of $(5, 0)$ and $(-1, 0)$:

$\left( \frac{5 + (-1)}{2}, \frac{0 + 0}{2} \right) = (2, 0)$

Therefore, the center of the hyperbola is $(h, k) = (2, 0)$.

The vertices are the points where the hyperbola intersects the transverse axis. Since the intercepts are $(5, 0)$ and $(-1, 0)$, these points serve as the vertices. The distance $a$ from the center to each vertex is:

$a = |5 - 2| = |-1 - 2| = 3$

Thus, $a = 3$, and $a^2 = 9$.

Using the Additional Points

The points $(2 \pm 3 \sqrt{2}, \pm 2)$ give us additional information to find $b^2$. The standard equation of the hyperbola with a horizontal transverse axis and center $(2, 0)$ is:

$\frac{(x-2)^2}{9} - \frac{y^2}{b^2} = 1$

Let's use the point $(2 + 3 \sqrt{2}, 2)$ to solve for $b^2$. Plugging these values into the equation:

$\frac{((2 + 3 \sqrt{2}) - 2)^2}{9} - \frac{(2)^2}{b^2} = 1$

$\frac{(3 \sqrt{2})^2}{9} - \frac{4}{b^2} = 1$

$\frac{18}{9} - \frac{4}{b^2} = 1$

$2 - \frac{4}{b^2} = 1$

$\frac{4}{b^2} = 1$

$b^2 = 4$

So, $b^2 = 4$.

The Equation of the Hyperbola

Now we have all the necessary components to write the equation of the hyperbola. We found that the center is $(2, 0)$, $a^2 = 9$, and $b^2 = 4$. The transverse axis is horizontal, so the equation is:

$\frac{(x-2)^2}{9} - \frac{y^2}{4} = 1$

This is the equation of the hyperbola that passes through the given points.

Graphing the Hyperbola

To graph the hyperbola, we need to identify the following elements:

1. Center: $(2, 0)$

2. Vertices: $(5, 0)$ and $(-1, 0)$

3. Asymptotes: To find the asymptotes, we use the equation:

   $y = \pm \frac{b}{a}(x - h) + k$

   In our case, $a = 3$, $b = 2$, $h = 2$, and $k = 0$. So the asymptotes are:

   $y = \pm \frac{2}{3}(x - 2)$

   Which gives us two asymptotes: $y = \frac{2}{3}(x - 2)$ and $y = -\frac{2}{3}(x - 2)$.

4. Additional Points: We can plot the given points $(2 \pm 3 \sqrt{2}, \pm 2)$ to further refine the graph.

The graph of the hyperbola will have two branches opening to the left and right, centered at $(2, 0)$. The vertices are at $(5, 0)$ and $(-1, 0)$, and the asymptotes act as guidelines for the branches as they extend away from the center.

Steps to Sketch the Graph

1. Plot the Center: Mark the point $(2, 0)$ on the coordinate plane.
2. Plot the Vertices: Mark the points $(5, 0)$ and $(-1, 0)$.
3. Draw the Asymptotes: Draw the lines $y = \frac{2}{3}(x - 2)$ and $y = -\frac{2}{3}(x - 2)$. These lines pass through the center and have slopes of $\pm \frac{2}{3}$.
4. Sketch the Hyperbola Branches: Starting from the vertices, sketch the two branches of the hyperbola, approaching the asymptotes as they extend away from the center. Ensure the branches pass through the given points $(2 \pm 3 \sqrt{2}, \pm 2)$.

Visual Representation

Describing a graph in text can be challenging. Ideally, this explanation would be accompanied by a visual graph. The hyperbola will be centered at $(2,0)$, open horizontally, and have asymptotes that intersect at the center, guiding the shape of the two branches. The vertices will be clearly visible on the x-axis, and the additional points will help in accurately sketching the curve.

Conclusion

We successfully derived the equation of the hyperbola, which is $\frac{(x-2)^2}{9} - \frac{y^2}{4} = 1$, using the given points. We also outlined the steps to sketch the graph of the hyperbola, emphasizing the importance of the center, vertices, and asymptotes. This exercise demonstrates the interplay between algebraic equations and geometric representations in understanding conic sections.

Understanding and graphing hyperbolas is a fundamental skill in analytic geometry. The process involves analyzing given points, determining key parameters such as the center, vertices, and asymptotes, and then using these parameters to formulate the equation and sketch the graph. This detailed approach provides a solid foundation for solving similar problems in conic sections.