Hyperbola Graph Features: Focus, Vertex, And Center

What's up, math whizzes! Today, we're diving deep into the fascinating world of conic sections, specifically hyperbole graphs. You know, those cool U-shaped curves that go on forever? We've got a challenge for you guys: we need to identify the key features of the hyperbola represented by the equation rac{y^2}{96^2}-rac{x^2}{40^2}=1. This isn't just about memorizing formulas; it's about understanding what these numbers mean visually on the graph. We need to pinpoint two correct descriptions from the given options, and trust me, once you get the hang of it, it's super satisfying to see how the equation translates directly to the shape and position of the hyperbola. So, grab your graphing paper (or just your brilliant minds!), and let's break down this equation piece by piece. We'll be looking at the center, vertices, and foci, and how they are derived from the standard form of a hyperbola equation. Get ready to flex those math muscles!

Understanding the Standard Equation of a Hyperbola

Alright guys, to nail down the features of our hyperbola, rac{y^2}{96^2}-rac{x^2}{40^2}=1, we first need to get cozy with its standard form. There are two main flavors of hyperbola equations, depending on whether the transverse axis is horizontal or vertical. Our equation fits the standard form for a vertical hyperbola:

$$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$$

Here's the lowdown on what $a$, $b$, and the center $(h,k)$ represent:

• Center $(h,k)$: This is the heart of the hyperbola, the point where the two axes of symmetry intersect. In our given equation, there are no $(x-h)$ or $(y-k)$ terms, which means $h=0$ and $k=0$. So, the center of this hyperbola is at $(0,0)$. This is a crucial starting point, guys, as all other features are measured relative to the center.

• $a^2$ and $b^2$: These denominators tell us about the 'spread' of the hyperbola along its axes. In the standard form rac{y^2}{a^2} - rac{x^2}{b^2} = 1, $a^2$ is the denominator under the positive term, and $b^2$ is under the negative term. For our equation, rac{y^2}{96^2}-rac{x^2}{40^2}=1:

  • $a^2 = 96^2$, which means $a = 96$. Since $a^2$ is under the $y^2$ term, the transverse axis is vertical. This means the hyperbola opens upwards and downwards.
  • $b^2 = 40^2$, which means $b = 40$. This value relates to the conjugate axis, which is horizontal in this case.

• Vertices: The vertices are the points where the hyperbola intersects its transverse axis. For a vertical hyperbola centered at $(0,0)$, the vertices are located at $(0, extbf{+}a)$ and $(0, extbf{-}a)$. Using our value $a=96$, the vertices are at $(0, 96)$ and $(0, -96)$.

• Foci: The foci are two fixed points that define the hyperbola. For a vertical hyperbola centered at $(0,0)$, the foci are located at $(0, extbf{+}c)$ and $(0, extbf{-}c)$, where $c^2 = a^2 + b^2$. Let's calculate $c$ for our hyperbola: $c^2 = 96^2 + 40^2 = 9216 + 1600 = 10816$. So, $c = \sqrt{10816}$. A quick calculation reveals $c = 104$. Therefore, the foci are at $(0, 104)$ and $(0, -104)$.

It's super important to remember the difference between $a$, $b$, and $c$. $a$ is always associated with the term that's positive in the equation and dictates the direction of the transverse axis. $b$ is associated with the negative term. And $c$ is always calculated using $c^2 = a^2 + b^2$ and tells us the distance to the foci from the center. Keep these relationships in mind, and you'll be able to decode any hyperbola equation like a pro!

Identifying the Vertices and Their Coordinates

Let's zero in on the vertices of our hyperbola, rac{y^2}{96^2}-rac{x^2}{40^2}=1. As we established, this is a vertical hyperbola because the $y^2$ term is positive. The standard form for a vertical hyperbola centered at $(h,k)$ is $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$. Since our equation has no shifts (i.e., $h=0$ and $k=0$), it's centered right at the origin $(0,0)$.

For a vertical hyperbola centered at the origin, the vertices are located at $(0, extbf{+}a)$ and $(0, extbf{-}a)$. From our equation, we can see that $a^2 = 96^2$, which means $a = 96$. Therefore, the vertices of this hyperbola are at $(0, 96)$ and $(0, -96)$.

Now, let's check our options:

• Option C suggests a vertex at $(-40,0)$. This would be a vertex for a horizontal hyperbola where $a=40$. But ours is vertical with $a=96$.
• Option D states a vertex at $(0,96)$. This matches one of our calculated vertices! So, this is a correct description.

It's easy to get mixed up between horizontal and vertical hyperbolas, guys. Remember, if $y^2$ is positive, it's a vertical hyperbola, and its vertices are on the y-axis. If $x^2$ is positive, it's a horizontal hyperbola, and its vertices are on the x-axis. The value of $a$ (the square root of the denominator of the positive term) always gives the distance from the center to the vertices along the transverse axis.

Locating the Foci of the Hyperbola

Moving on to the foci, these are super important points that define the unique shape of a hyperbola. For any hyperbola, the foci lie on the transverse axis, and their distance from the center is denoted by $c$. The relationship between $a$, $b$, and $c$ is key: $c^2 = a^2 + b^2$. This formula is different from the one for ellipses (where $c^2 = a^2 - b^2$ or $b^2 - a^2$), so pay close attention, guys!

For our vertical hyperbola, rac{y^2}{96^2}-rac{x^2}{40^2}=1, centered at $(0,0)$, the foci are located at $(0, extbf{+}c)$ and $(0, extbf{-}c)$. We already found $a=96$ and $b=40$. Let's plug these values into the formula for $c^2$:

$c^2 = a^2 + b^2 = 96^2 + 40^2$

$c^2 = 9216 + 1600$

$c^2 = 10816$

Now, we need to find $c$ by taking the square root of $10816$:

$c = \sqrt{10816}$

Calculating this, we find $c = 104$.

Since this is a vertical hyperbola, the foci are located at $(0, 104)$ and $(0, -104)$.

Let's check our options again:

• Option A suggests a focus at $(104,0)$. This would be a focus for a horizontal hyperbola where $c=104$. However, our hyperbola is vertical.
• Option B suggests a focus at $(0,-96)$. This coordinate is actually one of the vertices we found earlier, not a focus.

It seems like there might be a slight confusion in the options provided regarding the foci calculation. Based on our calculations, the foci are at $(0, 104)$ and $(0, -104)$. Let's re-examine option A: a focus at $(104,0)$. If the equation were $\frac{x^2}{96^2}-\frac{y^2}{40^2}=1$, then $a=96$ (horizontal transverse axis) and $c=104$, giving foci at $(\pm 104, 0)$. Our equation is $\frac{y^2}{96^2}-\frac{x^2}{40^2}=1$, which means the transverse axis is vertical, and the foci are on the y-axis. So, foci are $(0, \pm 104)$.

Let's assume there might be a typo in the options or the question implies that we should consider the magnitude of the distance $c$ from the center, and one of the coordinates of the foci might be presented in a way that requires careful interpretation or comparison. However, strictly following the standard forms, $(104,0)$ is not a focus for this vertical hyperbola. The foci are $(0, extbf{+}104)$.

Let's revisit the options one last time, ensuring we haven't missed anything subtle.

• A. a focus at $(104,0)$: Incorrect for a vertical hyperbola. Foci are on the transverse (y) axis.
• B. a focus at $(0,-96)$: Incorrect. $(0,-96)$ is a vertex.
• C. a vertex at $(-40,0)$: Incorrect. This would be a vertex for a horizontal hyperbola with $a=40$. Our vertices are on the y-axis.
• D. a vertex at $(0,96)$: Correct! This matches one of our calculated vertices.
• E. the center at $(0,0)$: Correct! As we determined from the equation's form.

It appears there was a mix-up in my initial assessment of the foci options. Let's clarify: the value $c=104$ is indeed the distance from the center to the foci. For our vertical hyperbola, the foci are at $(0, extbf{+}c)$, meaning $(0, 104)$ and $(0, -104)$. Option A, $(104,0)$, describes a focus for a horizontal hyperbola. Therefore, strictly speaking, it's not a feature of this graph. My apologies for any confusion, guys!

Determining the Center of the Hyperbola

The center of a hyperbola is arguably the most fundamental feature to identify, as it serves as the reference point for all other components like vertices and foci. The standard form of a hyperbola centered at $(h,k)$ is either $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ (horizontal transverse axis) or $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ (vertical transverse axis).

Our given equation is rac{y^2}{96^2}-rac{x^2}{40^2}=1. Notice that there are no terms like $(x-h)$ or $(y-k)$ present. This implies that $h=0$ and $k=0$. Therefore, the center of the hyperbola is at $(0,0)$. This is a direct consequence of the equation being in its simplest form, without any translation applied.

Let's look at Option E: