Problem 8
Question
Write the standard form of the equation of a hyperbola centered at \((h, k)\) that opens up and down.
Step-by-Step Solution
Verified Answer
The standard form is \( \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \).
1Step 1: Identify Form of Hyperbola
For a hyperbola centered at \((h, k)\) that opens up and down, the standard form of the equation is \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]. Here, \(a\) and \(b\) are real numbers such that \(a^2\) and \(b^2\) represent the squares of the distances from the center to the vertices and co-vertices respectively.
2Step 2: Substitute the Center
Since the problem specifies a hyperbola centered at \((h, k)\), substitute \(h\) and \(k\) directly into the equation format. The equation becomes \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \].
3Step 3: Explain Components
The numerator \((y - k)^2\) shows the vertical opening by centering around \(y = k\). The \(a^2\) under \((y - k)^2\) determines the length from the center to each vertex along the y-axis. Similarly, \((x - h)^2\) with \(b^2\) accounts for the x-axis distance to co-vertices, reflecting the transverse direction.
4Step 4: Final Equation Form
The hyperbola opens along the y-axis (up and down) as indicated by \((y - k)^2\) appearing first and positive. Thus, the equation is correctly stablished as \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \].
Key Concepts
Standard FormCenter of HyperbolaOpening of HyperbolaDistance to Vertices
Standard Form
When dealing with hyperbolas, the standard form of the equation is crucial for understanding their geometric properties. For a hyperbola that opens up and down, like the one given in the problem, the standard form is \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]. In this form, the expression \((y - k)^2\) is the leading component, reflecting vertical orientation.
It's important for keeping the hyperbola centered and correctly sized.
- \(a^2\) represents the vertical distance, squared, from the center to each vertex.
- \(b^2\) denotes the horizontal distance, squared, from the center to each co-vertex.
It's important for keeping the hyperbola centered and correctly sized.
Center of Hyperbola
The center of a hyperbola is akin to the core or middle point from where its shape stretches out. In our equation, the center is defined by the point \((h, k)\). This center point is not arbitrary—it's critical because:
- The coordinates \(h\) and \(k\) determine the position of the hyperbola on the graph.
- From this central point, the parts of the hyperbola expand upward and downward.
Opening of Hyperbola
The direction in which a hyperbola opens provides valuable information on its orientation in the coordinate plane. For the standard form equation \( \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \), this hyperbola opens vertically—up and down. Here's how we can tell:
- Since \((y - k)^2\) is the first term, the primary stretch occurs along the y-axis, revealing a vertical opening.
- The minus sign before the \((x - h)^2\) indicates that the x-direction contributes less to the shape's spread.
Distance to Vertices
The concept of vertices in a hyperbola involves understanding the stretch and expansion from its center. In our vertical hyperbola, these vertices are crucial because they illustrate the spread along the y-axis.
- The vertices are found \(a\) units up and down from the center \((h, k)\).
- The length from the center to a vertex is represented by \(a\), making the total distance between vertices \(2a\).
Other exercises in this chapter
Problem 7
Write the standard form of the equation of an ellipse centered at the origin and symmetric to both axes.
View solution Problem 8
Determine whether \((1,-1)\) is a solution of the system: $$\left\\{\begin{array}{l}2 x+y-1=0 \\ x^{2}-y^{2}=3\end{array}\right.$$
View solution Problem 8
Write the standard form of the equation of a horizontal or vertical ellipse centered at \((h, k)\)
View solution Problem 9
Find the solutions of the system \(\left\\{\begin{array}{l}x^{2}+4 y^{2}=25 \\\ x^{2}-2 y^{2}=1\end{array}\right.\) on the right.
View solution