Problem 13
Question
Write an equation of a parabola opening upward with a vertex at the origin. focus 1.5 units from vertex
Step-by-Step Solution
Verified Answer
The equation of the parabola opening upward with vertex at origin and focus 1.5 units from vertex is \(y = 4/3x^2\).
1Step 1: Identifying Given Information
In the given problem, we know that the vertex of the parabola is at the origin, so h=0 and k=0. Also, the parabola opens upward, which means the value of a will be positive. The focus is 1.5 units from the vertex, which can help us find the value of a.
2Step 2: Applying Formula of Focus to Find 'a'
The formula related to the focus of a parabola is \(a = 1/(4f)\), where f is the distance from the vertex to the focus. So substituting f=1.5 units, we can find a.
3Step 3: Substitute 'a' value in Equation
After calculation from Step 2, replace the calculated value of 'a' into the standard form equation \(y = a (x-h)^2 + k\). Since h and k are 0 (as the vertex is at the origin), the equation simplifies to y=ax^2.
4Step 4: Form the equation of Parabola
Once we have the value of 'a' and knowing that the vertex is at the origin, we can write down the equation of the parabola.
Key Concepts
Vertex FormFocus of a ParabolaUpward Opening Parabola
Vertex Form
The vertex form is a way of expressing the equation of a parabola. It showcases the vertex's position simply and effectively. A parabola's general equation in vertex form is expressed as:
\[ y = a(x-h)^2 + k \]
In this formula:
\[ y = a(x-h)^2 + k \]
In this formula:
- \( (h, k) \) represents the vertex's coordinates. It is essentially the "turning point" of the parabola.
- \( a \) dictates the parabola's width and direction of opening. If \( a \) is positive, the parabola opens upwards, while if \( a \) is negative, it opens downwards.
- Knowing the vertex is key to formulating an accurate equation for parabolic shapes. This format allows you to see shifts and shapes of the parabola relative to the vertex easily.
Focus of a Parabola
The focus of a parabola is a critical point that defines its shape and orientation. It lies along the axis of symmetry of the parabola.
For a parabola opening upwards, the focus is located above the vertex. This focus point is uniquely important because:
The mathematical formula used in our problem is \( a = \frac{1}{4f} \), where \( f \) is the distance from the vertex to the focus. This distance \( f \) helps detail how far the parabola opens relative to the focus.
For a parabola opening upwards, the focus is located above the vertex. This focus point is uniquely important because:
- Every point on the parabola is equidistant from the focus and a corresponding point on the "directrix," another line below the vertex.
- The formula connecting the focus to the parabola's equation helps calculate the parameter \( a \).
The mathematical formula used in our problem is \( a = \frac{1}{4f} \), where \( f \) is the distance from the vertex to the focus. This distance \( f \) helps detail how far the parabola opens relative to the focus.
Upward Opening Parabola
An upward opening parabola is characterized by its U-shape that opens away from the origin or the vertex.
This type of parabola means:
In general, these parabolas model many natural phenomena and are used in real-life applications like satellite dishes, water fountains, and even architecture design owing to their reflective properties.
This type of parabola means:
- The value of \( a \) is positive, indicating the parabola's arms extend upwards.
- The vertex is the lowest point on the graph when the parabola opens upwards.
In general, these parabolas model many natural phenomena and are used in real-life applications like satellite dishes, water fountains, and even architecture design owing to their reflective properties.
Other exercises in this chapter
Problem 13
Identify the conic section represented by each equation by writing the equation in standard form. For a parabola, give the vertex. For a circle, give the center
View solution Problem 13
Find the foci of each hyperbola. Then draw the graph. $$ \frac{x^{2}}{64}-\frac{y^{2}}{36}=1 $$
View solution Problem 13
Graph each equation. Identify the conic section and describe the graph and its lines of symmetry. Then find the domain and range. $$ x^{2}+9 y^{2}=1 $$
View solution Problem 13
Write an equation for each translation. $$ x^{2}+y^{2}=100 ; \text { down } 5 $$
View solution