Problem 32
Question
Find an equation for the conic that satisfies the given conditions. Parabola, vertex \((1,0), \quad\) directrix \(x=-5\)
Step-by-Step Solution
Verified Answer
The equation of the parabola is \(y^2 = 24(x - 1)\).
1Step 1: Identify the Orientation
Since the vertex of the parabola is at \((1,0)\) and the directrix is a vertical line \(x = -5\), the parabola opens to the right. This indicates it's a horizontal parabola.
2Step 2: Determine the Standard Form
The standard form of a horizontal parabola is \((y - k)^2 = 4p(x - h)\), where \((h,k)\) is the vertex of the parabola. Here, \(h=1\) and \(k=0\).
3Step 3: Calculate the Distance to the Directrix
The distance \(p\) from the vertex to the focus equals the distance from the vertex to the directrix. The vertex \((1,0)\) is \(6\) units away from the directrix \(x = -5\). Therefore, \(p = 6\).
4Step 4: Write the Equation
Substitute \(h = 1\), \(k = 0\), and \(p = 6\) into the standard form equation: \((y - 0)^2 = 4 \times 6 \times (x - 1)\). Simplify to get the equation \(y^2 = 24(x - 1)\).
Key Concepts
Parabola EquationVertex and DirectrixHorizontal Parabola
Parabola Equation
A parabola is one of the basic conic sections found in mathematics, characterized by its unique symmetrical U-shape. The equation of a parabola takes different forms depending on its orientation. For parabolas that open to the left or right, as in this example, we use the standard equation
The values of \(h\), \(k\), and \(p\) determine its position and orientation in the coordinate system. A positive value of \(p\) means the parabola extends to the right, while a negative \(p\) denoting a left-opening parabola. The concept becomes clearer by substituting these values appropriately.
- \((y - k)^2 = 4p(x - h)\)
- \((h, k)\) represents the vertex of the parabola.
- \(p\) is the distance from the vertex to the focus, a crucial detail for constructing the parabola.
The values of \(h\), \(k\), and \(p\) determine its position and orientation in the coordinate system. A positive value of \(p\) means the parabola extends to the right, while a negative \(p\) denoting a left-opening parabola. The concept becomes clearer by substituting these values appropriately.
Vertex and Directrix
The vertex and directrix are essential components in forming the parabola.
In this exercise, the vertex is given as
The directrix, on the other hand, is a fixed line used to define the parabola's "flattest" point in the geometric plane. Here, it is given by the vertical line
It serves to help construct and visualize the parabolic shape, as it sets a reference line opposite the focus. For any point on the parabola, the distance to the focus equals the distance to the directrix.
In this exercise, the vertex is given as
- \((1,0)\).
The directrix, on the other hand, is a fixed line used to define the parabola's "flattest" point in the geometric plane. Here, it is given by the vertical line
- \(x = -5\).
It serves to help construct and visualize the parabolic shape, as it sets a reference line opposite the focus. For any point on the parabola, the distance to the focus equals the distance to the directrix.
Horizontal Parabola
A horizontal parabola is a specific orientation where the parabola opens sideways instead of up or down. In our case, the nature of the given vertex and directrix tells us the parabola opens to the right.
When the vertex
The standard process involves determining key features:
When the vertex
- \((1,0)\)
- \(x = -5\)
The standard process involves determining key features:
- Direction: known from whether \(p\) is positive (right) or negative (left).
- Equation form: hinging on specific vertex-directrix distance.
- Length of \(p\): measuring distance from vertex to directrix, precisely 6 units here.
Other exercises in this chapter
Problem 31
\(29-48\) Sketch the curve with the given polar equation. $$r=\sin \theta$$
View solution Problem 31
(a) Show that the parametric equations $$x=x_{1}+\left(x_{2}-x_{1}\right) t \quad y=y_{1}+\left(y_{2}-y_{1}\right) t$$ where \(0 \leqslant t \leqslant 1,\) desc
View solution Problem 32
\(29-48\) Sketch the curve with the given polar equation. $$r=-3 \cos \theta$$
View solution Problem 33
Find an equation for the conic that satisfies the given conditions.\(y^{2}=-12(x+1)\) Parabola, focus \((-4,0), \quad\) directrix \(x=2\)
View solution