Problem 16 - Calculus Early Transcendentals

Question

Sketch a graph of the following parabolas. Specify the location of the focus and the equation of the directrix. Use a graphing utility to check your work. $$4 x=-y^{2}$$

Step-by-Step Solution

Answer

Answer: The given parabola has its vertex at the origin \((0,0)\) and opens to the left due to the negative coefficient of \(y^2\). It has a focus at \((-1,0)\) and the equation of the directrix is \(x=1\).

Step 1: Convert the equation into the standard form.

The given equation is \(4x=-y^2\). Divide both sides by \(4\), and we get: $$x = -\frac{1}{4}y^2$$ This is already in the standard form of a parabola with a vertical axis: $$x = \pm\frac{1}{4p}y^2$$ where \(p\) is the distance between the vertex and the focus or vertex and the directrix.

Step 2: Identify the value of p, axis, vertex, and orientation.

From the standard form we obtained, we have \(\frac{1}{4p} = -\frac{1}{4}\). Equating the coefficients, we find the value of \(p\): $$\frac{1}{4p} = -\frac{1}{4} \quad \Rightarrow \quad p=-1$$ Since the equation is in the form \(x=-\frac{1}{4}y^2\), the axis of the parabola is vertical, and it opens to the left as the coefficient of \(y^2\) is negative. The vertex of the parabola is at the origin \((0,0)\).

Step 3: Find the focus and the equation of the directrix.

Now that we have the value of \(p\), we can use it to find the focus and the equation of the directrix. Since the parabola opens to the left, the focus is \((-1,0)\) and the directrix is \(x=1\).

Step 4: Sketch the graph of the parabola.

To sketch the parabola, we will use its vertex, focus, and directrix. 1. Plot the vertex \((0,0)\). 2. Plot the focus \((-1,0)\). 3. Draw the vertical line \(x=1\) as the directrix. 4. Sketch the parabola symmetrically around the vertex and focus, ensuring that it's opening to the left and the points on the parabola are equidistant from the focus and directrix.

Step 5: Check the work using a graphing utility.

To confirm the sketch, use a graphing utility to plot the equation \(4x=-y^2\). The graph should match the sketch, showing the parabola with vertex at \((0,0)\), focus at \((-1,0)\), and the directrix at \(x=1\).

Key Concepts

Focus of a Parabola

The focus of a parabola is a special point that helps determine the shape and position of the curve. For a parabola defined by the equation \[ x = -\frac{1}{4}y^2 \],we find that the parabola is oriented vertically, with its focus given by the coordinates \((-1, 0)\).
This point is one unit to the left of the vertex at the origin, \((0,0)\).

But what makes this focus point so important? Every point on the parabola is equidistant from the focus and the directrix. This characteristic is what gives the parabola its U-shape.

• The location of the focus affects how "wide" or "narrow" the parabola appears.
• Moving the focus closer to the vertex creates a narrower parabola, while moving it further away results in a wider one.

Understanding where the focus lies can help you better sketch and understand the parabola's structure.

Directrix of a Parabola

The directrix of a parabola is not part of the curve itself, but an imaginary line that plays a crucial role. For the equation \[ x = -\frac{1}{4}y^2 \], the directrix is a vertical line with the equation \( x = 1 \).
This line is also one unit away from the vertex, but in the opposite direction to that of the focus.

Here's why the directrix matters:

• It helps maintain the relationship that each point on the parabola is equidistant to the focus and this line.
• The directrix, along with the focus, helps to define the set of points that create the parabolic curve.

Graphically, you can consider the directrix as a boundary.
The parabola will "hug" the focus and maintain an equal distance to the directrix at all points along its path.
Knowing the position of the directrix is essential for accurate graph sketching.

Vertex of a Parabola

The vertex of a parabola is the turning point and central feature of the parabola. For our specific equation, \[ 4x = -y^2 \],the vertex is conveniently located at the origin, \( (0, 0) \).
This point is crucial as it acts as the reference for both the focus and the directrix.

Several important aspects include:

• The vertex provides a balance point for the parabola.
• From this point, the parabola will mirror itself on either side, providing a symmetrical shape.
• The distance from the vertex to the focus gives us the parameter \( p \), which was calculated as \( p = -1 \) in our equation.

In essence, the vertex is where the axis of symmetry of the parabola cuts through, giving you a clear indication of the direction in which the parabola opens.
This central feature aids in constructing and interpreting the parabola's graph.