Problem 15 - Thomas Calculus

Question

Exercises \(9-16\) give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Include the focus and directrix in your sketch. $$ x=-3 y^{2} $$

Step-by-Step Solution

Answer

Focus: \((-\frac{3}{4}, 0)\), Directrix: \(x = \frac{3}{4}\). Parabola opens left.

Step 1: Identify the Parabola Equation

The given equation is \( x = -3y^2 \), which is a quadratic equation in terms of \( y \). Our goal is to identify the standard form of a parabola that opens horizontally. For a parabola opening left or right, the standard form is \((y - k)^2 = 4p(x - h)\). In this case, we will rewrite \( x = -3y^2 \) to fit this model.

Step 2: Rewrite Equation in Standard Form

Rewrite the equation \( x = -3y^2 \) into the form of \( (y - k)^2 = 4p(x - h) \). Here, \( y^2 = -\frac{1}{3}x \) represents \( (y - 0)^2 = -3(x + 0) \), implying \( k = 0 \), \( h = 0 \), and \( 4p = -3 \). Therefore, \( p = -\frac{3}{4} \).

Step 3: Identify Focus and Directrix

For the parabola \( (y - k)^2 = 4p(x - h) \), the vertex is at \((h, k)\). In this case, the vertex is \((0, 0)\). The focus is located at \((h + p, k)\), which gives us the focus at \((-\frac{3}{4}, 0)\). The directrix is the vertical line \( x = h - p \), thus \( x = \frac{3}{4} \).

Step 4: Sketch the Parabola

To sketch, plot the vertex at \((0, 0)\), the focus at \((-\frac{3}{4}, 0)\), and the directrix as the vertical line \( x = \frac{3}{4} \). Since \( p \) is negative, the parabola opens to the left (toward the focus). Draw the symmetrical parabola opening towards the focus, equidistant from the directrix.

Key Concepts

Parabola

A parabola is a unique set of points on a plane, forming a curve where each point is equidistant from a specific point and a line. This special point is known as the "focus" and the line is called the "directrix." Think of a parabola as the trajectory that a ball follows when it's thrown or the shape of satellite dishes.

• A parabola can open in various directions: upwards, downwards, leftwards, or rightwards, depending on its equation.
• The most basic orientation is defined by its coefficient: a positive coefficient means it opens up or to the right; a negative means down or to the left.

Each parabola has its axis of symmetry, which is a line that splits the parabola into two mirror-image halves. In this exercise, we have a horizontal parabola opening to the left.

Focus and Directrix

The focus and directrix are two crucial elements in defining a parabola. The focus is a point inside the parabola where all the reflected paths from the directrix converge. Meanwhile, the directrix is a straight line located outside of the parabola.

For the equation provided in the exercise, the focus is at the point \((h + p, k)\), and the directrix is represented by the vertical line \(x = h - p\). In our solved example, this translates to:

• The vertex of the parabola located at \((0, 0)\).
• The focus found at \((-rac{3}{4}, 0)\).
• The directrix being the line \(x = \frac{3}{4}\).

Understanding the focus and directrix allows you to graph the parabola accurately and visualize its path.

Standard Form of Parabola

The standard form of a parabola is critical for easily deducing its properties such as direction, vertex, and the position of the focus and directrix. For parabolas that open horizontally, the standard form is \[(y - k)^2 = 4p(x - h)\]

• The variable \(h\) and \(k\) denote coordinates of the vertex of the parabola.
• The term \(4p\) indicates the directional opening of the parabola; the value of \(p\) determines the distance from the vertex to the focus, and from the vertex to the directrix in the opposite direction.

In the problem's transformed equation, \((y - 0)^2 = -3(x + 0)\), it becomes apparent that:

• The parabola is centered at the origin \((0,0)\).
• It opens to the left as indicated by the negative coefficient of \(x\).
• The parameter \(p\) is calculated from \(4p = -3\), resulting in \(p = -\frac{3}{4}\).

A solid grasp of the standard form facilitates the transition from equation to graph, making the process smooth and intuitive.