Mastering Hyperbolas: Guide To $y^2-x^2+6x-4y-9=0$

Unraveling the Hyperbola Equation: From General Form to Standard Form

Hey guys, ever stared at a jumbled math equation like $y^2-x^2+6x-4y-9=0$ and wished it would just make sense? Well, today we're tackling one of those beasts, a hyperbola equation given in its messy general form. Our ultimate mission? To transform this tangled expression into a sleek, understandable standard form for a hyperbola. This foundational step is where all the magic begins, seriously! Without this transformation, trying to extract the hyperbola's key features would be like trying to navigate a city without a map—utterly confusing and inefficient. The general form gives you bits and pieces of information, but the standard form clearly lays out the hyperbola's center, its precise orientation, and its exact dimensions, which are all critical for understanding and graphing the curve. It's the absolute key to unlocking all its secrets, so let's get into it.

To begin, our strategy revolves around the powerful technique of completing the square. This method allows us to rewrite quadratic expressions into perfect square trinomials, which are essential for forming the $(x-h)^2$ and $(y-k)^2$ terms we need. Let's walk through it step-by-step with our equation: $y^2-x^2+6x-4y-9=0$.

First, we need to group the terms involving $x$ and $y$ together and move the constant term to the other side of the equation. This helps us organize our thoughts and set up the completing the square process:

$(y^2 - 4y) - (x^2 - 6x) = 9$

A quick but crucial note here, guys: when you factor out a negative sign from the $x$ terms, like we did with $-(x^2 - 6x)$, be super careful! The term inside the parenthesis changes sign, so $-x^2 + 6x$ becomes $-(x^2 - 6x)$. This is a common trip-up point, so always double-check your signs!

Next, let's complete the square for the $y$-terms. We have $y^2 - 4y$. To complete the square, we take half of the coefficient of our $y$ term (which is $-4$), square it, and add it. Half of $-4$ is $-2$, and $(-2)^2$ is $4$. So, we add $4$ inside the parenthesis for $y$. To keep the equation balanced, we must also add $4$ to the right side of the equation:

$(y^2 - 4y + 4) - (x^2 - 6x) = 9 + 4$

This simplifies to $(y-2)^2$.

Now, let's complete the square for the $x$-terms. We have $x^2 - 6x$. Half of the coefficient of our $x$ term (which is $-6$) is $-3$, and $(-3)^2$ is $9$. So, we add $9$ inside the parenthesis for $x$. However, remember that sneaky negative sign we factored out earlier? It means we're actually subtracting $9$ from the left side of the equation. Therefore, to maintain balance, we must also subtract $9$ from the right side of the equation:

$(y-2)^2 - (x^2 - 6x + 9) = 9 + 4 - 9$

This simplifies to $(x-3)^2$.

Combining these steps, our equation now looks like:

$(y-2)^2 - (x-3)^2 = 4$

We're almost there! The final step for the standard form is to make the right side of the equation equal to $1$. To achieve this, we divide every term in the equation by $4$:

rac{(y-2)^2}{4} - rac{(x-3)^2}{4} = rac{4}{4}

And voilà! Our beautiful standard form hyperbola equation is:

rac{(y-2)^2}{4} - rac{(x-3)^2}{4} = 1

This transformation is super important because it directly gives us h, k, a, and b—the foundational values for understanding our hyperbola. Without it, we'd be totally lost. It’s the foundational step, and getting it right sets the stage for everything else. Imagine trying to build a house without a proper foundation – disaster! So, mastering this "completing the square" technique is absolutely vital for understanding any conic section. We can clearly see now that our hyperbola has its y-term first, meaning it opens up and down, making its transverse axis vertical. This little detail already tells us a ton about its orientation even before we draw a single line. It's all about recognizing these patterns and understanding what each part of the equation signifies, allowing us to move confidently to the next steps of our analysis.

Pinpointing the Heart of the Hyperbola: Finding the Center

Alright, with our glorious hyperbola equation now in its standard form, rac{(y-2)^2}{4} - rac{(x-3)^2}{4} = 1, identifying the center of the hyperbola is an absolute breeze! This is probably one of the most straightforward steps, but don't let its simplicity fool you; it's profoundly important. Think of the center as the gravitational core, the unmoving point of perfect symmetry for the entire hyperbola. It's the anchor around which every other key feature—the vertices, the foci, and the asymptotes—is precisely and symmetrically arranged.

Remember the general standard form for a hyperbola. If the $y$-term is positive, indicating a vertical transverse axis (meaning the hyperbola opens up and down), the form is rac{(y-k)^2}{a^2} - rac{(x-h)^2}{b^2} = 1. If the $x$-term is positive, indicating a horizontal transverse axis, it's rac{(x-h)^2}{a^2} - rac{(y-k)^2}{b^2} = 1. Our equation clearly matches the first case, with the $(y-k)^2$ term coming first.

By direct comparison with our derived equation, rac{(y-2)^2}{4} - rac{(x-3)^2}{4} = 1, we can easily spot the values for $h$ and $k$. We see that $k=2$ and $h=3$. It's super important to remember that in the standard form $(x-h)$ and $(y-k)$, the signs are flipped when you pull out the coordinates. So, if it's $(x-3)$, the $h$ value is positive $3$. If it were $(x+3)$, then $h$ would be $-3$. Always keep an eye on those signs!

So, our center, the very heart of this hyperbola, is located at (3, 2).

What does this center point actually mean for our hyperbola? It's not a point on the curve itself (unlike a vertex), but it serves as the origin for the hyperbola's own internal coordinate system. Imagine drawing a faint set of axes through this point; everything else about the hyperbola will be measured and oriented relative to these new axes. If you were to conceptually fold the hyperbola in half, whether horizontally or vertically, the center would be your precise folding point. It's your steadfast reference point for every subsequent calculation and drawing step, a stable core from which the wild, sweeping branches of the hyperbola emerge.

Understanding and correctly identifying the center is absolutely critical for accurate graphing. It’s the very first point you’ll plot on your coordinate plane, and every other feature—the vertices, the foci, and the asymptotes—will radiate from there. Without a correctly identified center, your entire hyperbola will be shifted to the wrong location, and consequently, all your other calculations for the vertices, foci, and asymptotes will be fundamentally off. So, guys, always double-check your $(h, k)$ values! This step, while seemingly simple once you have the standard form, is truly the cornerstone for accurately visualizing and understanding the hyperbola's geometry and ensuring that your final graph is a true representation of the equation.

Mapping the Hyperbola's Key Features: Vertices and Foci Explained

Alright, team! With the hyperbola's center locked down at (3, 2), let's dive into some of the most exciting and defining points of our curve: the vertices and the foci. These points aren't just coordinates; they're the fundamental architects that dictate the hyperbola's shape, its spread, and its very mathematical definition. Understanding them is key to truly mastering this conic section.

First things first, let's identify our a and b values directly from our standard equation: rac{(y-2)^2}{4} - rac{(x-3)^2}{4} = 1. In this form, $a^2$ is always the denominator of the positive term, and $b^2$ is the denominator of the negative term.

• From the $(y-2)^2/4$ term, we have $a^2 = 4$, which means $a = extbf{2}$.
• From the $(x-3)^2/4$ term, we have $b^2 = 4$, which means $b = extbf{2}$.

What do these values tell us? The value of a is incredibly significant. It represents the distance from the center of the hyperbola to each of its vertices along the transverse axis. Since our $y$-term is positive in the standard form, our transverse axis is vertical. This means our hyperbola opens up and down, and the vertices will be directly above and below the center.

Let's find those vertices! These are the points where the hyperbola actually touches its transverse axis. They're like the