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Magazine Chapter 10: Problem 2
Find the vertex, focus, and directrix of the parabola and sketch its graph. $$4 y+x^{2}=0$$
Short Answer
Expert verified
Vertex: (0,0); Focus: (0,-1); Directrix: y=1.
Step by step solution
01
Rewrite the Equation in Standard Form
Start with the given equation: \(4y + x^2 = 0\). This equation is not in the standard form of a parabola yet. First, solve for \(y\) to express \(y\) as a function of \(x\): \[4y = -x^2\] Now, divide both sides by 4:\[y = -\frac{x^2}{4}\]. This equation represents a parabola in the standard form \(y = ax^2\) for horizontal opening.
02
Identify the Parabola Type
The equation \(y = -\frac{x^2}{4}\) indicates a parabola that opens downwards because the coefficient of \(x^2\) is negative. Compare this with the standard vertical equation \(y = ax^2\) where \(a = -\frac{1}{4}\). This identifies the parabola as vertically oriented, opening downwards.
03
Find the Vertex
For the equation \(y = ax^2\), the vertex form is given by \((h, k)\). Since the equation is already in the form \(y = -(1/4)x^2\), and there are no horizontal or vertical shifts, the vertex \((h, k)\) is at \((0,0)\).
04
Find the Focus and Directrix
For a parabola \(y = ax^2\), the focus is at \((0, \frac{1}{4a})\) and the directrix is given by \(y = -\frac{1}{4a}\). Here, \(a = -\frac{1}{4}\), so:- Focus: \(\left(0, \frac{1}{4\left(-1/4\right)}\right) = (0, -1)\)- Directrix: \(y = -\frac{1}{4\left(-1/4\right)} = y = 1\)
05
Sketch the Graph
The parabola opens downward with its vertex at the origin \((0,0)\). Mark the focus at \((0, -1)\) and draw the line for the directrix at \(y = 1\). The arms of the parabola will open downward since the coefficient of \(x^2\) is negative. Ensure your sketch reflects these characteristics: symmetrical about the \(y\)-axis, opening downward with focus and directrix positioned correctly.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vertex of a Parabola
The vertex of a parabola is a crucial concept when examining its properties and graphical representation. It represents the point where the parabola changes direction, offering a significant clue about the shape and orientation of the curve.
In the particular example of the equation given: \( y = -\frac{x^2}{4} \), the vertex is easily determined. The expression is in the simplest form \( y = ax^2 \), which places the vertex directly at the origin, \((0, 0)\).
In the particular example of the equation given: \( y = -\frac{x^2}{4} \), the vertex is easily determined. The expression is in the simplest form \( y = ax^2 \), which places the vertex directly at the origin, \((0, 0)\).
- For this equation type, if no additional linear or constant terms exist, the vertex is precisely at \((h, k) = (0, 0)\).
- The vertex indicates the turning point, meaning the curve will extend downward in this case, as suggested by the negative \(a\).
- Understanding the vertex assists in recognizing symmetry; it lies on the line of symmetry of the parabola, which is the \(y\)-axis for vertical parabolas.
Focus and Directrix
The focus and directrix of a parabola are geometric components that play an integral role in its definition. They provide information about the shape and depth of the parabola.
For the equation \( y = -\frac{x^2}{4} \), we use known formulas to locate these elements:
Also, remember that the parabola "opens" towards the focus. Here, the parabola opens downward, directed towards the focus at \((0, -1)\).
For the equation \( y = -\frac{x^2}{4} \), we use known formulas to locate these elements:
- Focus: The formula \((0, \frac{1}{4a})\) yields the position of the focus. Here, \(a = -\frac{1}{4}\), thus leading to the focus at \((0, -1)\).
- Directrix: This is a horizontal line calculated as \(y = -\frac{1}{4a}\). For our equation, it results in the line \(y = 1\).
Also, remember that the parabola "opens" towards the focus. Here, the parabola opens downward, directed towards the focus at \((0, -1)\).
Standard Form of a Parabola
A parabola's standard form is an algebraic representation that helps identify its type and structural properties instantly.
The standard forms differ for vertical and horizontal parabolas. The vertical parabola standard form, \( y = ax^2 + bx + c \), succinctly evolves in our example:
The standard forms differ for vertical and horizontal parabolas. The vertical parabola standard form, \( y = ax^2 + bx + c \), succinctly evolves in our example:
- The given equation, when rearranged to \( y = -\frac{x^2}{4} \), highlights the vertical orientation of the parabola with \(a = -\frac{1}{4}\).
- This form immediately tells us the parabola opens downward, as evidenced by entering a negative coefficient for \(x^2\).
- The vertex, position of focus, and directrix calculation stem from this form.
- Graphical characteristics such as width and directionality are determined by the \(a\) value鈥攄ictating the rate at which the parabola widens or narrows.
