Graphing $y^2 - 9x^2 = 1$: A Hyperbola Guide

Hey guys! Today, we're diving into the fascinating world of hyperbolas! Specifically, we're going to graph the equation $y^2 - 9x^2 = 1$. Don't worry, it's not as intimidating as it looks. We'll break it down step-by-step, making it super easy to understand. So, buckle up and let's get started!

1. Understanding the Hyperbola Equation

Before we jump into graphing, let's understand what the equation $y^2 - 9x^2 = 1$ represents. This is the standard form of a hyperbola centered at the origin (0,0). A hyperbola is a type of conic section, which basically means it's a curve formed by the intersection of a plane and a double cone. Think of two parabolas opening away from each other – that's essentially what a hyperbola looks like. In this equation, the $y^2$ term comes first and is positive, indicating that the hyperbola opens vertically. The general form of a hyperbola centered at the origin is either $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (opens horizontally) or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ (opens vertically). Our equation, $y^2 - 9x^2 = 1$, fits the second form. We can rewrite it as $\frac{y^2}{1^2} - \frac{x^2}{(1/3)^2} = 1$. This tells us that $a = 1$ and $b = \frac{1}{3}$. These values are crucial because 'a' determines the distance from the center to the vertices along the y-axis, and 'b' is related to the asymptotes of the hyperbola. The vertices are the points where the hyperbola intersects its main axis (in this case, the y-axis). The asymptotes are lines that the hyperbola approaches as it extends to infinity. They serve as guides for sketching the curve. Understanding the equation and its components is the foundation for accurately graphing the hyperbola. The orientation, vertices, and asymptotes are all derived from this equation, so make sure you're comfortable with these concepts before moving on. With these fundamentals in place, we can now proceed to isolate 'y' and prepare for the next graphing steps.

2. Solving for y: Isolating the Positive and Negative Square Roots

Okay, let's get our hands dirty with some algebra! Our goal here is to isolate 'y' in the equation $y^2 - 9x^2 = 1$. This will allow us to express 'y' as a function of 'x', which is what we need to graph it. First, we need to isolate the $y^2$ term. We can do this by adding $9x^2$ to both sides of the equation: $y^2 = 9x^2 + 1$. Now, to get 'y' by itself, we need to take the square root of both sides. Remember, when we take the square root, we need to consider both the positive and negative roots. This is super important because it gives us the two branches of the hyperbola. So, we get $y = \pm \sqrt{9x^2 + 1}$. This gives us two equations: $y = \sqrt{9x^2 + 1}$ and $y = -\sqrt{9x^2 + 1}$. The first equation, $y = \sqrt{9x^2 + 1}$, represents the upper branch of the hyperbola, which lies above the x-axis. The second equation, $y = -\sqrt{9x^2 + 1}$, represents the lower branch of the hyperbola, which lies below the x-axis. By graphing both of these equations, we will get the complete hyperbola. This separation into positive and negative roots is key to visualizing and accurately plotting the hyperbola. Make sure to keep track of which equation corresponds to which branch as you move on to the graphing stage. Understanding that the square root introduces both positive and negative solutions is a fundamental concept in algebra and is essential for graphing not only hyperbolas but also other types of equations. With 'y' now expressed in terms of 'x' for both branches, we are well-prepared to plot the graph.

3. Graphing the Two Equations

Alright, the moment we've been waiting for! Now we're going to graph the two equations we found: $y = \sqrt{9x^2 + 1}$ and $y = -\sqrt{9x^2 + 1}$. You can use a graphing calculator, an online graphing tool like Desmos or Geogebra, or even plot points by hand. If you're using a graphing calculator or online tool, simply enter the two equations and the tool will generate the graph for you. If you're plotting points by hand, choose a few values for 'x', like -2, -1, 0, 1, and 2. Plug these values into each equation to find the corresponding 'y' values. For example, if $x = 0$, then for $y = \sqrt{9x^2 + 1}$, we have $y = \sqrt{9(0)^2 + 1} = \sqrt{1} = 1$. And for $y = -\sqrt{9x^2 + 1}$, we have $y = -\sqrt{9(0)^2 + 1} = -\sqrt{1} = -1$. So, we have the points (0, 1) and (0, -1). These are actually the vertices of our hyperbola! Now, let's think about the asymptotes. For a hyperbola of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the asymptotes are given by the equations $y = \pm \frac{a}{b}x$. In our case, $a = 1$ and $b = \frac{1}{3}$, so the asymptotes are $y = \pm 3x$. These are the lines that the hyperbola approaches as 'x' gets very large or very small. Sketch these lines lightly on your graph. Now, plot the points you calculated and sketch the two branches of the hyperbola, making sure they approach the asymptotes as they extend outwards. The graph should look like two curves opening upwards and downwards, symmetric about the y-axis, and getting closer and closer to the lines $y = 3x$ and $y = -3x$. Remember, the hyperbola never actually touches the asymptotes, but they get incredibly close. With the vertices and asymptotes as guides, you should be able to sketch an accurate representation of the hyperbola $y^2 - 9x^2 = 1$.

4. Key Features of the Hyperbola

Let's recap the key features of the hyperbola $y^2 - 9x^2 = 1$ to solidify our understanding. Firstly, the center of the hyperbola is at the origin (0, 0). This is because there are no translations in the equation; it's in its simplest form. Secondly, the vertices are the points where the hyperbola intersects its main axis. In this case, the main axis is the y-axis, and the vertices are (0, 1) and (0, -1). These are the points we found earlier when we set x = 0. Thirdly, the asymptotes are the lines $y = 3x$ and $y = -3x$. These lines guide the shape of the hyperbola as it extends to infinity. The hyperbola approaches these lines but never actually touches them. Fourthly, the orientation of the hyperbola is vertical, meaning it opens upwards and downwards. This is because the $y^2$ term is positive in the equation. If the $x^2$ term were positive, the hyperbola would open horizontally. Fifthly, we can talk about the foci of the hyperbola, although we didn't explicitly calculate them during the graphing process. The foci are two points inside the hyperbola that are important in its definition. The distance between any point on the hyperbola and the two foci has a constant difference. The distance from the center to each focus is given by $c = \sqrt{a^2 + b^2}$. In our case, $a = 1$ and $b = \frac{1}{3}$, so $c = \sqrt{1 + \frac{1}{9}} = \sqrt{\frac{10}{9}} = \frac{\sqrt{10}}{3}$. Therefore, the foci are at the points $(0, \frac{\sqrt{10}}{3})$ and $(0, -\frac{\sqrt{10}}{3})$. Understanding these key features will not only help you graph hyperbolas more accurately but also allow you to analyze and interpret them in various contexts. Whether you're studying physics, engineering, or mathematics, hyperbolas pop up in many different applications, so having a solid grasp of their properties is essential.

5. Conclusion

So there you have it! We've successfully graphed the hyperbola $y^2 - 9x^2 = 1$ by solving for 'y' and graphing the two resulting equations. Remember, the key steps are: understanding the hyperbola equation, isolating 'y' (and remembering the positive and negative square roots!), graphing the two equations, and understanding the key features of the hyperbola. With practice, you'll become a hyperbola graphing pro in no time! Keep exploring and have fun with math!