Problem 16 - Contemporary Precalculus

Question

In Exercises \(11-16,\) find the focus and directrix of the parabola. $$x^{2}-3 y=0$$

Step-by-Step Solution

Answer

Question: Determine the focus and the directrix of the parabola given by the equation \(x^2-3y=0\). Answer: The focus of the given parabola is at the point \(F=(0,\dfrac{3}{4})\), and the equation of the directrix is \(y=-\dfrac{3}{4}\).

Step 1: 1. Rewrite the equation in the standard form

Move all terms to one side of the equation to form an equation of the form \(y=ax^2\). To do this, add \(3y\) to both sides: $$x^2-3y=0 \implies x^2=3y$$ Now we have \(y=\dfrac{1}{3}x^2\). This means \(a=\dfrac{1}{3}\).

Step 2: 2. Use the definition of a parabola

We can use the vertex form of a parabola, \(y=a(x-h)^2+k\), to help us determine the focus and directrix. For this equation, the vertex is located at the point (h, k). Since the parabola is given as \(y=\dfrac{1}{3}x^2\), the vertex is (h, k) = (0, 0). When the parabola opens upwards in the form \(y=ax^2\), the distance between the vertex and the focus is \(p = \dfrac{1}{4a}\). The equation of the directrix is \(y = -p\).

Step 3: 3. Calculate the coordinates of the focus and the directrix

The distance p is: $$p=\dfrac{1}{4a}=\dfrac{1}{4\cdot(1/3)}=\dfrac{1}{4\cdot\frac{1}{3}}=\dfrac{3}{4}$$ Since the parabola opens upwards and has the vertex at (0,0), the focus will be \(p\) distance upwards from the vertex. The coordinates of the focus (F) are: $$F=(h,k+p)=(0,0+\dfrac{3}{4})=(0,\dfrac{3}{4})$$ The equation of the directrix is \(y=k-p\): $$y = 0 - \dfrac{3}{4} = -\dfrac{3}{4}$$

Step 4: 4. Provide the answer

The focus of the parabola $$x^{2}-3 y=0$$ is $$F=(0,\dfrac{3}{4})$$ and the equation of the directrix is $$y=-\dfrac{3}{4}$$.

Key Concepts

Focus of a Parabola

The focus of a parabola is a crucial point which defines its geometric properties. In technical terms, the focus lies "inside" the parabola and is equidistant from all points on the curve along lines called "latus rectum". For our example, the equation of the parabola is given by \(x^2 - 3y = 0\).

• First, transform the equation into the standard form, which gives \(y = \frac{1}{3}x^2\).
• Here, \(a = \frac{1}{3}\), which helps in finding the position of the focus.

The focus is obtained as a point \((0, \frac{3}{4})\), which is above the vertex at a distance \(\frac{3}{4}\) along the y-axis. This point is essential because it defines the path of the parabola.

Directrix of a Parabola

The directrix of a parabola complements the focus by providing a line reference from which each point of the parabola is equidistant. For the parabola equation \(x^2 - 3y = 0\), the directrix formula applies based on distance \(p\).
Convert this equation into its standard form, \(y = \frac{1}{3}x^2\), to determine \(a = \frac{1}{3}\).

• This gives \(p = \frac{1}{4 \times \frac{1}{3}} = \frac{3}{4}\).

The directrix is then positioned at \(y = -\frac{3}{4}\). Observing this line, we can analyze how it is perpendicular to the axis of symmetry of the parabola, providing a complete understanding of its unique U shape.

Vertex Form of a Parabola

The vertex form of a parabola provides a simplified expression to determine the parabola's vertex, which serves as the "starting point" of the parabola. The standard vertex form is \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex.
In the example \(x^2 - 3y = 0\), the transformation to \(y = \frac{1}{3}x^2\) reveals:

• The vertex is at \((h, k) = (0, 0)\).
• This point is crucial as it determines the origin of the parabola in a coordinate plane.

This information is key to drawing the parabolic curve and understanding its symmetry and form in graphing scenarios.

Parabola Orientation

The orientation of a parabola refers to the direction in which the parabola opens. It is dictated by the leading coefficient \(a\) in the equation \(y = a(x-h)^2 + k\). For the equation \(y = \frac{1}{3}x^2\), since \(a = \frac{1}{3}\) is positive, the parabola opens upwards.
A few important points about parabola orientation include:

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), it opens downwards.
• If the standard equation were \(x = ay^2\), the orientation would be horizontal rather than vertical.

Understanding these orientations helps in identifying the direction and shape of the plotted curve, which is necessary for visualizing and solving geometric problems.