Problem 34
Question
Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Directrix has \(y\) -intercept 6
Step-by-Step Solution
Verified Answer
The equation is \(x^2 = 12y\).
1Step 1: Understanding parabolas
A parabola is the set of all points that are equidistant from a given point (focus) and a given line (directrix). In this problem, the vertex of the parabola is at the origin \((0,0)\).
2Step 2: Identify the condition
The directrix of the parabola is a horizontal line with the equation \(y = c\). Given the \(y\)-intercept of the directrix as 6, this means \(y = 6\).
3Step 3: Relation between focus and directrix
The vertex \((0, 0)\) is equidistant from the focus \((0,p)\) and the directrix \(y = 6\). Since these distances must match, we use the fact that \(2p = 6\) (because the distance from vertex to directrix is 6). Thus, \(p = 3\).
4Step 4: Write the equation of the parabola
For a parabola with vertex at the origin and directrix \(y = c\), the equation is \(x^2 = 4py\). Substituting \(p = 3\), we find \(x^2 = 12y\).
5Step 5: Verify the solution
The derived equation \(x^2 = 12y\) confirms that the parabola's focus \((0, 3)\) and directrix \(y = 6\) are equidistant from any point on the parabola.
Key Concepts
Vertex FormFocus and DirectrixEquidistant Points
Vertex Form
The vertex form of a parabola's equation is a powerful tool that helps us understand its shape and position easily. The vertex is the point where the parabola changes direction, also known as the "turning point." In vertex form, the equation of a parabola is written as \[ y = a(x-h)^2 + k \] where \((h, k)\) are the coordinates of the vertex.
- "a" determines the parabola's width and whether it opens upwards or downwards. If "a" is positive, the parabola opens upwards; if negative, it opens downwards.
- For instance, in our exercise, since the vertex is at the origin \((0,0)\), the terms "+h" and "+k" vanish, simplifying the equation to \( y = ax^2 \) when expressed in vertex form.
Focus and Directrix
The terms "focus" and "directrix" are essential in understanding the specific position and orientation of a parabola.
The focus is a point inside the parabola, whereas the directrix is a line outside of it. Every point on a parabola is equidistant from the focus and the directrix.
The focus is a point inside the parabola, whereas the directrix is a line outside of it. Every point on a parabola is equidistant from the focus and the directrix.
- In our specific problem, the directrix is a horizontal line given by the equation \( y = 6 \). So, it runs parallel to the x-axis, located 6 units above the origin.
- The focus, found at \((0, p)\) for parabolas with vertex at origin, is critical because it affects the "tightness" and direction in which the parabola opens.
- We found "p" by ensuring the distance from the vertex to the directrix matches the vertex to the focus: the calculation for "p" involved averaging the distances, leading to a focus at \((0, 3)\).
Equidistant Points
The concept of equidistant points is foundational in plotting a parabola. A parabola is defined as the set of all points that are equidistant from the focus and the directrix. This property ensures that any point on the parabola can be traced back to this balance between these two crucial components.
- When we say a point is equidistant, we mean that the perpendicular distance from the point to the directrix is the same as the distance from the point to the focus.
- In our parabola example, the point \((x, y)\) on the parabola will fulfill the condition: distance to the line \( y=6 \) equals the distance to the point \((0, 3)\).
- This property is why the derived equation \(x^2 = 12y\) effectively captures all points on the parabola.
Other exercises in this chapter
Problem 34
Use a graphing device to graph the conic. $$ x^{2}-4 y^{2}+4 x+8 y=0 $$
View solution Problem 34
Find an equation for the hyperbola that satisfies the given conditions. Vertices \((0, \pm 6),\) hyperbola passes through \((-5,9)\)
View solution Problem 35
If a projectile is fired with an initial speed of \(v_{0} \mathrm{ft} / \mathrm{s}\) at an angle \(\alpha\) above the horizontal, then its position after \(t\)
View solution Problem 35
Find an equation for the ellipse that satisfies the given conditions. Endpoints of major axis \(( \pm 10,0),\) distance between foci 6
View solution