Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the coordinates of the focus and the equation of the directrix for each parabola. Make a sketch showing the parabola, its focus, and its directrix. $$6 y-2 x^{2}=0$$

Step-by-Step Solution

Answer

Focus: (0, 3/4); Directrix: y = -3/4.

Step 1: Rewrite the Equation

First, rewrite the given equation in the standard form of a parabola that opens vertically. The given equation is: \( 6y - 2x^2 = 0 \).Rearrange the equation to isolate \( y \) on one side:\( 6y = 2x^2 \)Divide everything by 6 to simplify:\( y = \frac{1}{3}x^2 \)

Step 2: Identify Parabola Parameters

The standard form of a parabola that opens vertically is \( y = ax^2 \). Comparing to our equation \( y = \frac{1}{3}x^2 \), we identify that \( a = \frac{1}{3} \).

Step 3: Determine the Focus

The focus of a parabola \( y = ax^2 \) is located at \( \left(0, \frac{1}{4a}\right) \).Substitute \( a = \frac{1}{3} \) into the formula:\( \frac{1}{4a} = \frac{1}{(4 \times \frac{1}{3})} = \frac{3}{4} \).Thus, the coordinates of the focus are \( (0, \frac{3}{4}) \).

Step 4: Find the Equation of the Directrix

The equation of the directrix for a parabola \( y = ax^2 \) is given by \( y = -\frac{1}{4a} \).Using \( a = \frac{1}{3} \), calculate:\( y = -\frac{1}{4 \times \frac{1}{3}} = -\frac{3}{4} \).Thus, the equation of the directrix is \( y = -\frac{3}{4} \).

Step 5: Sketch the Parabola

On a coordinate plane, plot the vertex of the parabola at the origin \((0,0)\), since the parabola is \( y = \frac{1}{3}x^2 \). Add the focus \((0, \frac{3}{4})\) above the vertex and draw the directrix line at \( y = -\frac{3}{4} \).Sketch the U-shaped curve that opens upwards, ensuring it extends equally on both sides of the y-axis, passing through the vertex.

Key Concepts

Focus of a Parabola

The focus of a parabola is a key point that defines the shape and direction of the parabola itself. In the standard form of a vertically opening parabola, which is given by the equation \( y = ax^2 \), the focus can be found using the formula \( \left(0, \frac{1}{4a}\right) \). This focus point is located along the axis of symmetry, perpendicular to the directrix.

• The focus is always inside the parabola.
• It helps guide the "U" shape of the curve.
• The distance from the vertex to the focus is equal to the distance from the vertex to the directrix.

In our example, the equation \( y = \frac{1}{3}x^2 \) places the focus at \( (0, \frac{3}{4}) \). The fact that the focus is above the vertex indicates that the parabola opens upwards. Knowing how to find the focus is crucial for sketching the curve accurately, as it ensures a proper understanding of how far the parabola extends.

Directrix

The directrix is a straight line associated with a parabola and is vital for defining its geometry. It is located opposite the focus in relation to the vertex. For a vertical parabola with the equation \( y = ax^2 \), the directrix can be calculated using the formula \( y = -\frac{1}{4a} \). This line provides a boundary that the parabola "bends away" from.

• The directrix helps in determining the width and orientation of the parabola.
• It is exactly the same distance from the vertex as the focus is, but in the opposite direction.
• It does not intersect with the parabola itself.

In our exercise, substituting \( a = \frac{1}{3} \) into the formula, we find the directrix at \( y = -\frac{3}{4} \). This means it lies below the vertex of the parabola. Understanding the role of the directrix is essential for constructing accurate graphs, as it shows how the parabola curves away and helps define the overall shape.

Vertex Form

The vertex form of a parabola is a way of expressing the quadratic equation that makes its features easier to discern. It is given by the formula \( y = a(x - h)^2 + k \), where \( (h, k) \) represents the vertex coordinates. However, when the vertex is at the origin, \( (0, 0) \), the equation simplifies to \( y = ax^2 \), like in our problem.

• The vertex form is extremely useful for instantly identifying the vertex of the parabola.
• It helps in quickly determining direction (upward or downward) based on the sign of \( a \).
• It allows for easy calculation of important points such as the focus and the position of the directrix.

In the exercise, \( y = \frac{1}{3}x^2 \) reveals that the parabola opens upwards with its vertex at the origin \( (0, 0) \). The value of \( a = \frac{1}{3} \) indicates a vertically stretched parabola that is wider than the basic \( y = x^2 \).Being familiar with the vertex form gives you a solid foundation for analyzing and sketching parabolas effectively, as it outlines the core components and positioning with precision.