Problem 2 - Calculus Early Transcendentals

Question

Find the vertex, focus, and directrix of the parabola and sketch its graph. $$4 y+x^{2}=0$$

Step-by-Step Solution

Answer

Vertex: (0,0); Focus: (0,-1); Directrix: y=1.

Step 1: 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.

Step 2: 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.

Step 3: 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 \((0,0)\).

Step 4: 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: \(y = -\frac{1}{4\left(-1/4\right)} = y = 1\)

Step 5: 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.

Key Concepts

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, \((0, 0)\).

• 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:

• 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 \(y = 1\).

The focus and the directrix maintain a constant distance from any point on the parabola, ensuring the parabola's perfect geometric symmetry.
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 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 \(x^2\).

Recognizing the standard form helps in visualizing how the parabola behaves:

• 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—dictating the rate at which the parabola widens or narrows.

Hence, achieving understanding of the standard form provides the foundational instruments for predicting and sketching the parabola's graph accurately.