Problem 44
Question
Vertex \((0,0)\), focus \(\left(0,-\frac{7}{2}\right) \quad x^{2}=-14 y\)
Step-by-Step Solution
Verified Answer
Vertex is (0,0) and focus is (0,-\frac{7}{2}). The equation correctly describes the parabola.
1Step 1: Understanding the Equation
The given equation is \(x^{2} = -14y\), which is a standard form of a parabola that opens downwards. The general formula for a vertical parabola is \(x^{2} = 4py\), where \(p\) is the distance from the vertex to the focus.
2Step 2: Finding the Value of p
To find \(p\), compare the equation \(x^{2} = -14y\) with the standard form \(x^{2} = 4py\). Here, we have \(4p = -14\). Solving for \(p\), we get: \[ p = \frac{-14}{4} = -\frac{7}{2} \].
3Step 3: Locating the Focus
Since the vertex is at \((0, 0)\) and \(p = -\frac{7}{2}\), the focus is located \(p\) units below the vertex along the y-axis because the parabola opens downwards. Therefore, the focus is at \((0, -\frac{7}{2})\).
4Step 4: Verifying the Focus
According to the problem, the focus is at \((0, -\frac{7}{2})\) which matches with our calculated position based on the value of \(p\). Therefore, the identified focus is correct.
5Step 5: Conclusion
The equation \(x^{2} = -14y\) correctly describes a parabola with vertex \((0,0)\) and focus \((0, -\frac{7}{2})\) as given and analyzed.
Key Concepts
ParabolasVertexFocusStandard Form Equation
Parabolas
Parabolas are curved shapes that have a unique set of characteristics. If you imagine slicing a cone at just the right angle, the shape you get is a parabola. These curves are commonly seen in quadratic functions and are represented on a graph. They can either open upwards, downwards, leftwards, or rightwards, depending on the equation involved.
Here are a few important facts about parabolas:
Here are a few important facts about parabolas:
- They have a singular curve that continuously rises or falls without any loops.
- The path of a thrown ball or the trajectory of an arrow often resembles a parabola.
- Mathematically, when given as a quadratic function in the form of \( y = ax^2 + bx + c \), you're dealing with a parabola.
Vertex
The vertex of a parabola is a special point where the curve changes direction. In simpler terms, it's either the lowest or highest point of the parabola, depending on whether it opens upwards or downwards. For the parabola given in the example, which is \( x^2 = -14y \), the vertex is at (0,0).
Key characteristics of the vertex include:
Key characteristics of the vertex include:
- It acts as the turning point of the parabola.
- For a parabola described by \( y = ax^2 + bx + c \), the vertex can be found using the formula \( x = \frac{-b}{2a} \) for the x-coordinate, while substituting this value back into the equation gives the y-coordinate.
- In a vertical parabola, it serves as a point equidistant to the focus and the directrix.
Focus
The focus of a parabola is another defining point located inside the curve. It is not on the parabola itself but is vital in determining the shape of the curve. In the exercise given, the focus is at \(\left(0, -\frac{7}{2}\right)\).
Here's what makes the focus important:
Here's what makes the focus important:
- It lies directly in line with the vertex along the axis of symmetry.
- The points on the parabola are equidistant to the focus and a line called the directrix.
- In the case of a downward-opening parabola, like \( x^2 = -14y \), the focus is found by moving down from the vertex by \( p \) units, where \( p \) is the distance from vertex to focus.
Standard Form Equation
The standard form of a parabola's equation provides a simple yet comprehensive description of its geometry. For parabolas that open vertically, the equation is usually written as \( x^2 = 4py \), where \( p \) gives the distance from the vertex to the focus.
This form is particularly useful because:
This form is particularly useful because:
- It easily shows whether the parabola opens up or down; a positive \( p \) means it opens upwards, while a negative \( p \) indicates a downward opening, as seen in \( x^2 = -14y \).
- It provides immediate insight into the parabola's axis of symmetry and vertex placement.
- By setting specific points like a vertex and focus, you can derive the standard form tailored to a particular parabola.
Other exercises in this chapter
Problem 42
Find the equation of the circle that passes through the three points \((3,0),(6,-9)\) and \((10,-1)\).
View solution Problem 43
What is the graph of the equation \(x^{2}+y^{2}=0\) ? Explain your answer.
View solution Problem 44
What is the graph of the equation \(x^{2}+y^{2}=-4\) ? \(\mathrm{Ex}-\) plain your answer.
View solution Problem 45
What type of figure is the graph of the equation \(x^{2}+\) \(6 x+2 y^{2}-20 y+59=0 ?\) Explain your answer.
View solution