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: \((-\frac{1}{12}, 0)\); Directrix: \(x = \frac{1}{12}\).

Step 1: Identify the Standard Form of the Equation

The given equation is \( x = -3y^2 \). This is similar to the standard form of a parabola given by \( x = ay^2 \) where the parabola opens to the left or right depending on the sign of \( a \). Since \( a = -3 \), the parabola opens to the left.

Step 2: Determine the Vertex

For parabolas of the form \( x = ay^2 \), the vertex is at the origin \((0,0)\). Thus, for the given equation \( x = -3y^2 \), the vertex of the parabola is at \((0,0)\).

Step 3: Find the Focus and Directrix

The formula for the distance from the vertex to the focus for a parabola of the form \( x = ay^2 \) is \( \,\frac{1}{4a} \,\). For \( a = -3 \), the distance is \( \,\frac{1}{4(-3)} = -\frac{1}{12} \,\). The negative sign indicates the focus is to the left of the vertex by \( \frac{1}{12} \) units, so the focus is at \( \left(-\frac{1}{12}, 0\right)\). The directrix is a vertical line on the opposite side of the vertex from the focus, located at \( x = \frac{1}{12} \).

Step 4: Sketch the Parabola

Plot the vertex at \((0,0)\) on a coordinate plane. Mark the focus at \( \left(-\frac{1}{12}, 0\right)\). Then, draw the vertical line \( x = \frac{1}{12} \) as the directrix. Sketch a parabola that opens to the left with the vertex at the origin, equidistant from the focus and the directrix.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is one of its most important characteristics, serving as the point at which the curve changes direction. Consider it the 'tip' or 'turning point' of the parabola.
To find the vertex for a parabola in the form of \( x = ay^2 \), it's often located at the origin, that is,

• (0, 0), if there's no horizontal or vertical shift.

In our problem, the equation \( x = -3y^2 \) is already expressed in this standard form, revealing that the vertex is indeed at the origin, (0, 0). Being familiar with this type of equation allows us to swiftly determine the vertex, without needing further calculations.

Focus of a Parabola

The focus of a parabola plays a crucial role in its geometric definition. It's an internal point where all the 'reflected' lines or 'incoming' lines travel through. For a parabola in the form \( x = ay^2 \):

• The focus is located along the axis of symmetry of the parabola.
• The distance from the vertex to the focus is \( \frac{1}{4a} \).

In this scenario, with \( a = -3 \), the calculation becomes \( \frac{1}{4(-3)} = -\frac{1}{12} \). The negative sign indicates that the focus lies to the left of the vertex along the x-axis. Consequently, the focus for our parabola at \( x = -3y^2 \) is positioned precisely at \( \left(-\frac{1}{12}, 0\right) \). This demonstrates how the negative sign in the \( x = ay^2 \) equation affects the focus's position.

Directrix of a Parabola

While the focus tells us about a point inside the parabola, the directrix complements it by representing a line outside of it. The directrix is an invisible boundary that helps define the curve:

• For a horizontal opening parabola like \( x = ay^2 \), the directrix is a vertical line.
• It lies equidistant from the vertex as the focus but on the opposite side.

Given that our parabola's focus is \( \left(-\frac{1}{12}, 0\right) \), the directrix must be at \( x = \frac{1}{12} \). This indicates it's the same distance from the vertex as the focus, illustrating an essential geometric principle that defines every point on the parabola as being equidistant to both the focus and the directrix.