Problem 48
Question
Given information about the graph of the hyperbola, find its equation. Center: \((0,0) ;\) vertex: \((0,-13) ;\) one focus: \((0, \sqrt{313})\)
Step-by-Step Solution
Verified Answer
The equation of the hyperbola is \( \frac{y^2}{169} - \frac{x^2}{144} = 1 \).
1Step 1: Identify the Hyperbola Type
Since the center of the hyperbola is at the origin \(0, 0\), and the vertex is along the y-axis \(0, -13\), it indicates that the hyperbola is vertical. The general form for a vertical hyperbola is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]
2Step 2: Find the Value of \(a\)
The distance from the center to the vertex of the hyperbola is the value of \(a\). Given the vertex at \(0, -13\), we have \(a = 13\). This means \(a^2 = 169\).
3Step 3: Determine the Relationship Involving \(c\)
The distance from the center to a focus is \(c\). Here, one focus is at \(0, \sqrt{313}\), giving us \(c = \sqrt{313}\).
4Step 4: Use the Relationship \(c^2 = a^2 + b^2\)
Using the relationship, calculate \(b^2\):\[ c^2 = a^2 + b^2 \]\[ 313 = 169 + b^2 \]\[ b^2 = 313 - 169 = 144 \]
5Step 5: Write the Equation of the Hyperbola
Substitute the values of \(a^2\) and \(b^2\) into the general equation:\[ \frac{y^2}{169} - \frac{x^2}{144} = 1 \]
Key Concepts
Vertical HyperbolaConic SectionsCenter of HyperbolaVertexFocus Point
Vertical Hyperbola
A vertical hyperbola is a type of conic section where the two branches open upwards and downwards along the y-axis. This can be visualized as two mirror image curves that bow away from each other. The general equation for a vertical hyperbola centered at the origin is given by:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]In this equation:
- \(y^2\) : represents the vertical orientation, as 'y' comes first.
- \(a^2\) : is the distance from the center to the vertices along the y-axis.
- \(b^2\) : contributes to the width of the branches along the x-axis.
Conic Sections
Conic sections are curves obtained by intersecting a plane with a cone. They come in four types: circles, ellipses, parabolas, and hyperbolas. Hyperbolas appear when the intersecting plane cuts through both nappes of the cone.
Here's why they matter:
- Each conic section has unique properties, symmetry, and equations.
- Understanding conic sections is critical for physics, engineering, and astronomy.
Center of Hyperbola
The center of a hyperbola is a key point from which the geometry of the hyperbola is established. For hyperbolas centered at the origin, the center is at \((0, 0)\). From this point, distances are measured to determine other elements such as vertices and foci.Key points to remember:
- The center in our example is the starting point for defining the hyperbola's equation.
- It acts as the coordinate origin for symmetry.
- Knowing the center helps in graphing the hyperbola and understanding its orientation.
Vertex
The vertex of a hyperbola is like the tip of its curve when looking at one of the branches. In a vertical hyperbola, vertices are aligned along the y-axis. Here's what makes vertices special:
- Vertices are found at a distance \(a\) from the center.
- They help determine the shape and size of the hyperbola.
Focus Point
Focus points, or foci, are crucial to defining a hyperbola's shape and position. Each hyperbola has two foci that mirror the center and establish the spacing of the branches outward.Things to note:
- In the given problem, one focus is at \((0, \sqrt{313})\).
- Distance from the center to a focus is denoted as \(c\).
- The difference in distances to the foci remains constant along any tangent line to the hyperbola.
Other exercises in this chapter
Problem 48
For the following exercises, given information about the graph of the hyperbola, find its equation. Center: (0,0) ; vertex: (0,-13) ; one focus: \((0, \sqrt{313
View solution Problem 48
For the following exercises, graph the equation relative to the \(x^{\prime} y^{\prime}\) system in which the equation has no \(x^{\prime} y^{\prime}\) 'term. $
View solution Problem 49
For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=-\frac{1}{4}
View solution Problem 49
Find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Directrix: \(x=-\frac{1}{4} ; e=\frac{7}{2}\)
View solution