Problem 6 - Algebra and Trigonometry

Question

Find the vertex, focus, and directrix of the parabola, and sketch the graph. $$ (y+5)^{2}=-6 x+12 $$

Step-by-Step Solution

Answer

Vertex: (2, -5); Focus: (1/2, -5); Directrix: x = 7/2.

Step 1: Rewrite the Parabola Equation

Start by rewriting the given equation \((y + 5)^2 = -6x + 12\) in standard parabola form. The equation of a parabola with its axis parallel to the x-axis is typically given by \((y - k)^2 = 4p(x - h)\). First, isolate the \(x\) term: \((y + 5)^2 = -6x + 12\) to \((y + 5)^2 = -6(x - 2)\). Now, it resembles the form \((y - k)^2 = 4p(x - h)\).

Step 2: Identify Parameters for the Parabola

In the equation \((y + 5)^2 = -6(x - 2)\), compare it with \((y - k)^2 = 4p(x - h)\) to find \(k = -5\), \(h = 2\), and \(4p = -6\). Therefore, \(p = -\frac{3}{2}\).

Step 3: Find the Vertex of the Parabola

Since the standard form for a horizontal parabola \((y - k)^2 = 4p(x - h)\) has its vertex at \((h, k)\), the vertex here is \((2, -5)\).

Step 4: Find the Focus of the Parabola

The focus of a horizontal parabola is found using the formula \((h + p, k)\). Substituting the values, the focus is at \(\left(2 - \frac{3}{2}, -5\right) = \left(\frac{1}{2}, -5\right)\).

Step 5: Determine the Directrix of the Parabola

The directrix of a horizontal parabola is a vertical line given by the formula \(x = h - p\). Thus, the directrix is \(x = 2 + \frac{3}{2} = \frac{7}{2}\).

Step 6: Sketch the Graph of the Parabola

To sketch the parabola, plot the vertex at \((2, -5)\). The focus \((\frac{1}{2}, -5)\) and the directrix \(x = \frac{7}{2}\) help illustrate the shape. Indicate that the parabola opens to the left (since \(p\) is negative). Draw the symmetric curve around the axis passing through the focus and perpendicular to the directrix.

Key Concepts

Vertex

The vertex is a key point in a parabola and can be viewed as the tip or turning point of the curve. For horizontal parabolas, the vertex acts as the middle point between the focus and the directrix. In the standard form of a horizontal parabola equation

• \((y - k)^2 = 4p(x - h)\)

the vertex is located at the coordinates

• \((h, k)\)

. The vertex represents the core from which the parabola stretches out sideways. In the provided equation,

• \((y + 5)^2 = -6(x - 2)\)

the vertex is calculated by identifying the values

• \(h = 2\) and \(k = -5\)

, resulting in the vertex being at the point

• \((2, -5)\)

. This point provides a reference location from which the behavior of the parabola is determined, showing the center of its symmetrical properties.

Focus

The focus of a parabola is an internal point that, together with the directrix, defines its structure and orientation. This point is critical because it dictates how the parabola stretches and curves. For a horizontal parabola, the focus is typically given by the formula

• \((h + p, k)\)

where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus. For the equation

• \((y + 5)^2 = -6(x - 2)\)

we calculated

• \(p = -\frac{3}{2}\)

. Plugging in these values:

• The focus is found at \(\left(2 - \frac{3}{2}, -5\right)\)

which simplifies to

• \(\left(\frac{1}{2}, -5\right)\)

. This focus indicates the direction in which the parabola opens (to the left, in this case) and influences the geometry of the curve.

Directrix

A parabola's directrix is a straight line that, together with the focus, aids in defining its structure. Located opposite the stretch of the parabola from the focus, the directrix plays a key role in the parabola's reflectivity property, ensuring that any point on the parabola is equidistant from both the focus and the directrix. For a horizontal parabola, the directrix is expressed as

• \(x = h - p\)

In the equation given,

• \((y + 5)^2 = -6(x - 2)\)

the directrix is determined by the previously calculated

• \(p = -\frac{3}{2}\)

Therefore, the directrix is:

• \(x = 2 + \frac{3}{2} = \frac{7}{2}\)

. It is a vertical line at \(x = \frac{7}{2}\), and provides a crucial boundary that helps shape the trajectory of the parabola as it encases the focus.