Hyperbola Equation: Vertex (0,-8), Focus (0,10)

Hey guys, let's dive into the wild world of conic sections and tackle a cool problem about hyperbolas! We're on a mission to find the specific equation that represents a hyperbola centered at the origin, with a vertex chilling at $(0,-8)$ and a focus hanging out at $(0,10)$. This ain't just about memorizing formulas, but understanding the why behind them. So, grab your thinking caps, and let's break it down.

Understanding Hyperbola Basics

First off, what exactly is a hyperbola? Imagine two identical cones joined at their tips. A hyperbola is what you get when you slice through these cones with a plane at a specific angle. It's characterized by two separate, mirror-image branches that extend infinitely. For our problem, the hyperbola is centered at the origin, which is $(0,0)$. This simplifies things a lot because we're dealing with the standard forms of the hyperbola equations. Remember, guys, the standard form for a hyperbola centered at the origin looks a bit different depending on whether it opens vertically or horizontally.

If the hyperbola opens vertically (up and down), the equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Notice how the $y^2$ term is positive. If it opens horizontally (left and right), the equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, with the $x^2$ term being positive. The key players here are '$a$' and '$b$'. '$a$' represents the distance from the center to the vertices, and '$b$' is related to the conjugate axis and helps define the shape and width of the hyperbola. We also have the foci, which are points inside each branch that are crucial for defining the hyperbola's shape. The distance from the center to each focus is denoted by '$c$'. A super important relationship that binds $a$, $b$, and $c$ in a hyperbola is $c^2 = a^2 + b^2$. This is different from an ellipse, where it's $c^2 = a^2 - b^2$ (or $b^2 - a^2$). So, keep that $c^2 = a^2 + b^2$ in your back pocket!

Decoding the Vertex and Focus

Now, let's look at the juicy details given in our problem: a vertex at $(0,-8)$ and a focus at $(0,10)$. Since both the vertex and the focus lie on the y-axis (the x-coordinate is 0), this tells us our hyperbola is vertical. This is a major clue, guys! It means we'll be using the standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. The vertex is the point on the hyperbola closest to the center. For a vertical hyperbola centered at the origin, the vertices are at $(0, a)$ and $(0, -a)$. Our given vertex is $(0,-8)$. By comparing $(0,-8)$ with $(0,-a)$, we can immediately deduce that $a = 8$. Since $a$ represents a distance, it's always positive, so $a=8$ is correct. This also means $a^2 = 8^2 = 64$.

Next, let's talk about the focus. The foci of a vertical hyperbola centered at the origin are located at $(0, c)$ and $(0, -c)$. We are given a focus at $(0,10)$. This means $c = 10$. Again, $c$ is a distance, so it's positive. Therefore, $c^2 = 10^2 = 100$. We now have the values for $a^2$ and $c^2$. We can plug these into our trusty relationship $c^2 = a^2 + b^2$ to find $b^2$.

Calculating 'b^2' and Finalizing the Equation

We know that $c^2 = 100$ and $a^2 = 64$. Let's substitute these values into the equation $c^2 = a^2 + b^2$:

$100 = 64 + b^2$

To find $b^2$, we just need to subtract 64 from both sides:

$b^2 = 100 - 64$

$b^2 = 36$

Awesome! We've found $b^2$. Now we have all the components we need to write the equation of our hyperbola. We determined earlier that because the vertex and focus are on the y-axis, the hyperbola is vertical, and its equation is of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. We found $a^2 = 64$ and $b^2 = 36$. Plugging these values in, we get:

$\frac{y^2}{64} - \frac{x^2}{36} = 1$

This is the equation that perfectly represents the hyperbola described in the problem. It's centered at the origin, has a vertical orientation, a vertex at $(0,-8)$, and its foci will be at $(0, \pm c)$, where $c = \sqrt{a^2+b^2} = \sqrt{64+36} = \sqrt{100} = 10$. So, the foci are at $(0, 10)$ and $(0, -10)$. This matches all the conditions given!

Analyzing the Options

Let's quickly check this against the provided options to make sure we're on the right track:

A. $\frac{x^2}{64}-\frac{y^2}{36}=1$: This represents a horizontal hyperbola since the $x^2$ term is positive. Our vertex and focus are on the y-axis, so this is incorrect.

B. $\frac{x^2}{64}-\frac{y^2}{100}=1$: Another horizontal hyperbola. Also incorrect.

C. $\frac{y^2}{64}-\frac{x^2}{36}=1$: This is a vertical hyperbola. Here, $a^2 = 64$ (so $a=8$) and $b^2 = 36$. This matches exactly what we found! The vertices would be at $(0, \pm 8)$ and the foci would be at $(0, \pm \sqrt{64+36}) = (0, \pm 10)$. This is our winning ticket, guys!

D. This is listed as 'Discussion category : mathematics', which isn't a valid equation. So, definitely not this one!

Key Takeaways for Hyperbolas

So, what did we learn from this adventure, folks?

1. Identify the Orientation: The location of the vertex and focus is your biggest clue. If they lie on the y-axis, it's a vertical hyperbola ($\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$). If they lie on the x-axis, it's a horizontal hyperbola ($\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$).
2. Find 'a' and 'c': For a vertical hyperbola centered at the origin, vertices are $(0, \pm a)$ and foci are $(0, \pm c)$. For a horizontal one, vertices are $(\pm a, 0)$ and foci are $(\pm c, 0)$. The absolute values of the non-zero coordinates give you $a$ and $c$.
3. Use the Magic Formula: The relationship between $a$, $b$, and $c$ for hyperbolas is $c^2 = a^2 + b^2$. This is crucial for finding the missing piece, $b^2$.
4. Plug and Play: Once you have $a^2$ and $b^2$, plug them into the correct standard form equation based on the hyperbola's orientation.

Understanding these steps makes solving hyperbola problems a breeze. It's all about connecting the given information (vertex, focus) to the standard forms and using the fundamental relationships between $a$, $b$, and $c$. Keep practicing, and you'll be a hyperbola pro in no time! Remember, math is like a puzzle, and each piece of information helps you see the bigger picture. Keep those brains sharp!