Find an equation for the hyperbola described. Graph the equation by hand.

Center at​ (0,0); focus at

​(00​,1010​);

vertex at

​(00​,negative 6−6​)

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Part 1

An equation of the hyperbola is

enter your response hereminus−enter your response hereequals=1

​(Use integers or fractions for any numbers in the​ equation.)

Find an equation for the ellipse.  Graph the equation.

center at​ (0,0); focus at

​(0,2424​);

vertex at

​(0,negative 25−25​)

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Part 1

Type the left side of the equation of the ellipse.

enter your response hereequals=1

### Problem 1: Equation of the Hyperbola
Given:  
- Center: \((0, 0)\)  
- Focus: \((0, \pm 10)\)  
- Vertex: \((0, \pm 6)\)  

#### Step 1: Hyperbola Orientation and Standard Form
The hyperbola opens vertically because both the focus and vertex are along the \(y\)-axis. The standard form of such a hyperbola is:  
\[
\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
\]  
Here:  
- \(a\) is the distance from the center to a vertex.  
- \(c\) is the distance from the center to a focus.  
- The relationship \(c^2 = a^2 + b^2\) holds for hyperbolas.

#### Step 2: Determine \(a^2\), \(c^2\), and \(b^2\)
- \(a = 6\), so \(a^2 = 36\).  
- \(c = 10\), so \(c^2 = 100\).  
Using \(c^2 = a^2 + b^2\):  
\[
100 = 36 + b^2 \implies b^2 = 64.
\]  

#### Step 3: Equation of the Hyperbola
Substitute \(a^2\) and \(b^2\) into the standard form:  
\[
\frac{y^2}{36} - \frac{x^2}{64} = 1.
\]  

### Problem 2: Equation of the Ellipse
Given:  
- Center: \((0, 0)\)  
- Focus: \((0, \pm 24)\)  
- Vertex: \((0, \pm 25)\)  

#### Step 1: Ellipse Orientation and Standard Form
The ellipse is vertical, so the standard form is:  
\[
\frac{y^2}{a^2} + \frac{x^2}{b^2} = 1
\]  
Here:  
- \(a\) is the distance from the center to a vertex.  
- \(c\) is the distance from the center to a focus.  
- The relationship \(c^2 = a^2 - b^2\) holds for ellipses.

#### Step 2: Determine \(a^2\), \(c^2\), and \(b^2\)
- \(a = 25\), so \(a^2 = 625\).  
- \(c = 24\), so \(c^2 = 576\).  
Using \(c^2 = a^2 - b^2\):  
\[
576 = 625 - b^2 \implies b^2 = 49.
\]  

#### Step 3: Equation of the Ellipse
Substitute \(a^2\) and \(b^2\) into the standard form:  
\[
\frac{y^2}{625} + \frac{x^2}{49} = 1.
\]  

---

### Final Equations
1. **Hyperbola:**  
\[
\frac{y^2}{36} - \frac{x^2}{64} = 1.
\]

2. **Ellipse:**  
\[
\frac{y^2}{625} + \frac{x^2}{49} = 1.
\]  

Would you like further assistance with graphing these equations?  

---

### Related Questions:
1. What is the geometric significance of \(a\), \(b\), and \(c\) in hyperbolas and ellipses?  
2. How do you differentiate between the equations of ellipses and hyperbolas?  
3. What steps are involved in graphing a hyperbola by hand?  
4. How do the foci influence the shape of an ellipse or hyperbola?  
5. What are the asymptotes of the hyperbola, and how are they calculated?  

**Tip:** For hyperbolas, always determine the slopes of the asymptotes—they guide accurate graphing!

Find an equation for the hyperbola described. Graph the equation by hand. Center at (0,0); focus at (0, ±10); vertex at (0, ±6). Find an equation for the ellipse. Graph the equation. Center at (0,0); focus at (0, ±24); vertex at (0, ±25).

Learn to derive and graph equations for hyperbolas and ellipses centered at (0,0). Solve problems with given foci and vertices using standard forms and relationships between a, b, and c. Suitable for high school students in Grades 10-12.

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