Problem 27
Question
Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Focus \(F(-8,0)\)
Step-by-Step Solution
Verified Answer
The equation is \(y^2 = -32x\).
1Step 1: Identify the Parabola Type
Since the vertex of the parabola is at the origin, \(0,0\), and the focus, \((-8,0)\), lies on the x-axis, the parabola opens horizontally (left or right).
2Step 2: Equation of the Parabola
The standard form for a parabola that opens horizontally with its vertex at the origin is \(y^2 = 4px\). Here, \(p\) is the distance from the vertex to the focus.
3Step 3: Determine the Value of p
Since the focus is at \((-8,0)\), and the vertex at \(0,0\), \(p = -8\) (the negative sign indicates the parabola opens left).
4Step 4: Formulate the Equation
Substituting \(p = -8\) into the equation \(y^2 = 4px\), gives \[y^2 = 4(-8)x = -32x.\Therefore, the equation of the parabola is \space y^2 = -32x.\]
Key Concepts
Vertex FormFocus of a ParabolaOpening Direction of ParabolaDistance Calculation in Parabolas
Vertex Form
The vertex form of a parabola is a particular way of expressing the parabola's equation, which directly provides information about its vertex. The vertex is a crucial point that represents the "tip" or turning point of the parabola. For parabolas that open horizontally, the general vertex form looks like this:
- For horizontal parabolas: \[ y^2 = 4p(x - h) + k\]
- For vertical parabolas: \[ x^2 = 4p(y - k) + h\]
Focus of a Parabola
The focus of a parabola is a special point that, together with the directrix (a line), defines the parabola. Each point on the parabola is equidistant from the focus and the directrix. This characteristic forms the basis of the parabola's geometric definition.
In the problem at hand, the focus is given as \(F(-8, 0)\). With the vertex at the origin \(0,0\), the focus helps us not only to understand the shape of the parabola but also its orientation. The distance between the vertex and the focus is denoted by \(p\), which in this case is -8. The sign of \(p\) indicates the direction in which the parabola opens. Recognizing the focus and its placement relative to the vertex is crucial in graphing the parabola and understanding its physical shape.
In the problem at hand, the focus is given as \(F(-8, 0)\). With the vertex at the origin \(0,0\), the focus helps us not only to understand the shape of the parabola but also its orientation. The distance between the vertex and the focus is denoted by \(p\), which in this case is -8. The sign of \(p\) indicates the direction in which the parabola opens. Recognizing the focus and its placement relative to the vertex is crucial in graphing the parabola and understanding its physical shape.
Opening Direction of Parabola
The direction in which a parabola opens is vital in understanding its behavior. A parabola can open upwards, downwards, left, or right. When given the vertex at the origin and a specific focus point, determining this direction becomes straightforward.In our example, since the focus \(F(-8, 0)\) is located on the negative x-axis, and the vertex is at \(0,0\), the parabola opens to the left. If the parabola opens to the left or right, we know that it is a horizontal parabola, and its equation takes the form \(y^2 = 4px\).
- If \(p > 0\), the parabola opens to the right.
- If \(p < 0\), it opens to the left.
Distance Calculation in Parabolas
Calculating distances in parabolas is fundamental for constructing their equations and understanding their configurations. The parameter \(p\) represents the distance from the vertex of the parabola to its focus.In this specific problem, the given focus is \(-8,0\), while the vertex is at the origin \(0,0\). The distance \(p\) is effectively the x-coordinate of the focus, meaning \(p = -8\). The negative sign shows direction on the x-axis.The distance calculation between these two points forms the crucial link in determining \(p\) for the parabola's equation \(y^2 = 4px\). Once you know \(p\), you can substitute it directly into the parabola's equation form to find the complete equation. Understanding this distance calculation enables accurate formulation of the parabola's key characteristics.
Other exercises in this chapter
Problem 27
Find an equation for the hyperbola that satisfies the given conditions. Foci \((\pm 5,0),\) vertices \((\pm 3,0)\)
View solution Problem 27
Use a graphing device to graph the ellipse. $$6 x^{2}+y^{2}=36$$
View solution Problem 28
Find parametric equations for the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$
View solution Problem 28
(a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$x^{2}-2 x y+3 y^{2}=8$$
View solution