Problem 5
Question
Write an equation of a hyperbola with the given characteristics. vertices \((1,-3)\) and \((-7,-3),\) foci \((2,-3)\) and \((-8,-3)\)
Step-by-Step Solution
Verified Answer
After revising Step 2, the equation of the hyperbola is \((x+3)^{2}/16 - (y+3)^{2}/25 =1\). This is a horizontal hyperbola, as \(a^{2}\) is under the \(x\) term.
1Step 1: Find the center of the hyperbola
The center of the hyperbola is the midpoint of both the vertices and the foci. By calculating the average of the x-coordinates and the y-coordinates of the vertices (or foci), \((1 + -7)/2 = -3\) and \((-3+ -3)/2 = -3\), we find the center to be (-3,-3).
2Step 2: Determine the value of a
The value of \(a\) is the distance from the center to a vertex. Since our vertices are (1,-3) and (-7,-3), and our center is (-3,-3), we calculate \(a\) as |-3+3|=0.
3Step 3: Determine the value of c
The value of \(c\) is the distance from the center to a focus. Given the foci (2,-3) and (-8,-3), we calculate \(c\) as |-3+8|=5.
4Step 4: Determine the value of b
The relationship between \(a\), \(b\), and \(c\) in a hyperbola is \(c^{2}=a^{2}+b^{2}\). We can rearrange the formula to solve for \(b\) : \(b=\sqrt{c^{2}-a^{2}} = \sqrt{5^{2}-0^{2}} = 5\).
5Step 5: Write the equation of the hyperbola
The equation for a horizontal hyperbola is \((x-h)^{2}/a^{2} - (y-k)^{2}/b^{2}=1\), where (h, k) is the center of the hyperbola. Plugging in the values, we find the equation to be \((x+3)^{2}/0^{2} - (y+3)^{2}/25 =1\). However, as \(a^{2} = 0\), this equation doesn't exist in real numbers as we can't divide by zero. This means we've made a mistake at Step 2: the value of \(a\) should be distances from a vertex to the center, |-3+1|=4.
Key Concepts
VerticesFociCenter of HyperbolaEquation of Hyperbola
Vertices
In the context of a hyperbola, vertices are two distinct points that lie on the hyperbola and are closest to the center of the hyperbola. They are essentially the starting points of each branch of the hyperbola.
A hyperbola will have two vertices, and in this exercise, these vertices are given as
A hyperbola will have two vertices, and in this exercise, these vertices are given as
- (1, -3) and (-7, -3).
Foci
The foci of a hyperbola are key points that help define the steepness and orientation of its branches. Each hyperbola has two foci, and in this exercise, those foci are
The distance from the center of the hyperbola to each focus is known as 'c'. This distance helps describe the eccentricity of the hyperbola, which explains how "stretched" the shape is. In this exercise, calculating this distance (
- (2, -3) and (-8, -3).
The distance from the center of the hyperbola to each focus is known as 'c'. This distance helps describe the eccentricity of the hyperbola, which explains how "stretched" the shape is. In this exercise, calculating this distance (
c = 5) is critical for solving the hyperbola equation. Foci are important because points on the hyperbola's arms are such that the difference between the distances to the two foci is constant.Center of Hyperbola
The center of a hyperbola is a crucial point from which key properties of the hyperbola are measured. It lies exactly halfway between each pair of vertices and each pair of foci.
For the given exercise, by calculating the midpoint of the vertices and the foci, we determine the center at (-3, -3).
For the given exercise, by calculating the midpoint of the vertices and the foci, we determine the center at (-3, -3).
- Calculate the midpoint between the vertices or foci to find the center: For vertices i.e., midpoint \(\left(\frac{1 + (-7)}{2},\frac{-3 + (-3)}{2}\right) = (-3, -3)\).
a) and foci (c) are made. If any calculations involving the center are inaccurate, the resulting equation and graph of the hyperbola will also be inaccurate.Equation of Hyperbola
The equation of a hyperbola represents the mathematical relationship between all points that make up the hyperbola. It can appear intimidating due to its fraction-based form, but once you break it down, it becomes much clearer.
In the exercise, we are working with a horizontal hyperbola because the x-coordinates of the vertices change while the y-coordinates remain constant.
The standard form for a horizontal hyperbola is:
When completing your equation, be mindful of the squared terms and what they represent.
After correcting the error in calculating
In the exercise, we are working with a horizontal hyperbola because the x-coordinates of the vertices change while the y-coordinates remain constant.
The standard form for a horizontal hyperbola is:
- \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\].
(h, k) is the center, a is the distance from the center to a vertex, and b is determined through \(\sqrt{c^2 - a^2}\).When completing your equation, be mindful of the squared terms and what they represent.
After correcting the error in calculating
a in the exercise (by measuring from the center to the vertices properly), the equation was shown to be influenced directly by correct determination of a. Always double-check these calculations to ensure the hyperbola’s parameters are accurate.Other exercises in this chapter
Problem 4
Write an equation of a circle with the given center and radius. Check your answers. $$ (-6,10), 1 $$
View solution Problem 5
Write an equation of an ellipse in standard form with center at the origin and with the given vertex and co-vertex. $$ (0,-7),(4,0) $$
View solution Problem 5
Graph each equation. $$ 36 y^{2}-9 x^{2}=324 $$
View solution Problem 5
Write an equation for a graph that is the set of all points in the plane that are equidistant from the given point and the given line. $$ F(0,4), y=0 $$
View solution