Problem 53
Question
Find the standard form of the equation of the parabola with the given characteristics. Vertex: (0,2)\(;\) directrix: \(y=4\)
Step-by-Step Solution
Verified Answer
The standard form of the equation of the parabola is \(y=-\frac{1}{8}x^2+2\).
1Step 1: Find the Vertex
The vertex is given as (0,2). Therefore, h=0 and k=2.
2Step 2: Calculate the distance
The directrix is given as \(y=4\). To calculate 'p', find the distance between the vertex and the directrix. In this case, \(p=k-4 = 2-4 = -2\).
3Step 3: Find 'a'
We can now find 'a' using the equation \(a=\frac{1}{4p}\). Substituting the value of 'p' we found into the formula will give \(a=\frac{1}{4*(-2)}=-\frac{1}{8}\).
4Step 4: Formulate the Equation
Substitute h, k, and a into the parabola equation. Therefore, the equation of the parabola is \(y=-\frac{1}{8}(x-0)^2+2\) or simplified \(y=-\frac{1}{8}x^2+2\).
Key Concepts
Vertex FormDirectrixDistance to DirectrixStandard Form of a Parabola
Vertex Form
The vertex form of a parabola is a particularly neat way to express the equation of a parabola. This form makes it easy to locate the vertex of the parabola directly from the equation. It is generally represented as \[ y = a(x-h)^2 + k \] where
- \( (h, k) \) represents the vertex.
- \( a \) indicates the parabola's direction and width.
Directrix
The directrix is a fixed line used in combination with the focus to define a parabola. The unique feature of a parabola is that any point on it is equidistant from the focus and the directrix. In this exercise, the directrix is given as \(y = 4\). The directrix influences the direction and also the position of the vertex relative to the rest of the parabola. For vertical parabolas like the one in our exercise, the directrix is horizontal \((y=k+p)\), and it assists in confirming the shape's orientation. By knowing the directrix, we ascertain how the parabola is shaped and the inherent symmetry around its axis.
Distance to Directrix
Understanding the distance to the directrix is vital because this distance determines the parameter \(p\), which is crucial for other calculations. In our exercise, the given vertex is at \((0, 2)\) and the directrix at \(y = 4\). To find \(p\), calculate the distance between these two points, which results in \[ p = k - ext{directrix} = 2 - 4 = -2 \]The value of \(p\) helps define the orientation and opening direction of the parabola. A positive \(p\) would indicate a parabola opening upwards, while a negative value suggests it opens downwards. The absolute value indicates the distance from the vertex to the directrix, reflecting the parabolic curve's steepness and virality.
Standard Form of a Parabola
The standard form of a parabola is a common algebraic representation that highlights symmetry. In a basic scenario where a parabola opens vertically, the equation is of the form: \[ y = ax^2 + bx + c \]The standard form derived from our exercise is \[ y = -\frac{1}{8}x^2 + 2 \]highlighting a downward-opening parabola due to the negative \(a\) value. This form was rearranged from the vertex form with already known \(a\) as \(-\frac{1}{8}\). The standard form is useful for analyzing the parabola's intercepts and overall progression along the coordinate plane. Each term—the quadratic \(ax^2\), linear \(bx\), and constant \(c\)—plays a role, particularly in deriving roots and determining the exact path the parabola follows on a graph.
Other exercises in this chapter
Problem 53
Find the angle \(\theta\) (in radians and degrees) between the lines. $$\begin{aligned} &0.05 x-0.03 y=0.21\\\ &0.07 x+0.02 y=0.16 \end{aligned}$$
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Sketch (if possible) the graph of the degenerate conic. $$x^{2}+y^{2}+2 x-4 y+5=0$$
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Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$9 x^{2}+4 y^{2}-18 x+16 y-119=0$$
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Find an equation of the ellipse with vertices (±5,0) and eccentricity \(e=\frac{3}{5}.\)
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