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. $$12 x=5 y^{2}$$

Short answer

Answer: The parabola opens horizontally to the right. The coordinates of the focus are (3/5, 0), and the equation of the directrix is x = -3/5.

Step-by-step solution

Rewrite the equation in the standard form

To do this, we will solve the equation for y in terms of x: $$12x = 5y^2$$ $$y^2 = \frac{12}{5}x$$ Now the equation is in the form $$y^2 = 4px$$, where $$4p = \frac{12}{5}$$.

Identify the orientation of the parabola

Since the equation is in the form of $$y^2 = 4px$$, the parabola opens horizontally to the right.

Find the vertex

The vertex of the parabola is at the origin (0, 0) since there are no translations in x or y.

Calculate the value of p

To find the value of p, we can set $$4p = \frac{12}{5}$$ and solve for p: $$p = \frac{12}{5} \div 4 = \frac{12}{5} \cdot \frac{1}{4} = \frac{12}{20} = \frac{3}{5}$$

Determine the focus and directrix

Since the parabola opens to the right, we will add the value of p to the x-coordinate of the vertex to find the focus, and subtract p from the x-coordinate of the vertex to find the directrix: Focus: $$(\frac{3}{5}, 0)$$ Directrix: $$x = -\frac{3}{5}$$

Sketch the graph

Now we can sketch the graph using the information we have gathered: 1. Plot the vertex at the origin (0, 0). 2. Plot the focus at the point $$(\frac{3}{5}, 0)$$. 3. Draw the directrix as a vertical line at $$x = -\frac{3}{5}$$. 4. Sketch the parabola opening to the right, with the vertex at the origin, and passing through the focus. Finally, use a graphing utility to check the sketch. The plotted parabola should resemble the graph for the equation $$y^2 = \frac{12}{5}x$$, and the focus and directrix locations should match the calculated values.

Key concepts

Focus and Directrix of Parabola

When dealing with parabolas, the focus and directrix are essential components defining its shape. Every parabola is uniquely characterized by its focus—a specific point—and its directrix—a line.

• The focus is a point to which every point on the parabola is equidistant from the directrix, which means the parabola "leans towards" its focus.
• The directrix is a line perpendicular to the axis of symmetry of the parabola. It lies on the opposite side of the vertex from the focus.

In the equation \(12x = 5y^2\), by rewriting it as \(y^2 = \frac{12}{5}x\), you identify that the focus is at \(\left(\frac{3}{5}, 0\right)\) and the directrix is a vertical line at \(x = - \frac{3}{5}\).
The distance \(p\), calculated as \(\frac{3}{5}\), measures the span from the vertex to the focus and from the vertex to the directrix equally. This means the vertex sits exactly halfway between the focus and directrix.

Parabola Orientation

Parabolas can open in four different ways: upward, downward, to the left, or to the right. In the case of the parabola described by \(y^2 = \frac{12}{5}x\), the distinct feature is that the \(y^2\) term suggests the parabola opens horizontally.

• If the equation is in the form \(y^2 = 4px\), the parabola opens to the right when \(p > 0\).
• Conversely, it opens to the left if \(p < 0\).

Here, because \(p = \frac{3}{5} > 0\), the parabola indeed opens to the right. Recognizing the orientation is useful as it helps you to correctly position and sketch the parabola and anticipate its shape and direction.

Parabola Vertex

The vertex of a parabola is its turning point and the central point of symmetry. You find the vertex by identifying any translations from the origin as described in the equation.

• For the equation \(y^2 = \frac{12}{5}x\), there are no constant terms or linear terms, indicating no horizontal or vertical translations from the origin.
• This places the vertex precisely at the origin \((0, 0)\), given the absence of added or subtracted numbers in the equation.

The vertex serves as an anchor point for both the focus and directrix calculations. It’s the point from which the parabola symmetrically expands outwards towards its focus and away from its directrix. Knowing the vertex’s location helps establish the rest of the parabola's features.