Problem 41 - Algebra and Trigonometry

Question

In Exercises 35–42, find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$ (y+1)^{2}=-8 x $$

Step-by-Step Solution

Answer

The vertex of the parabola is at point (0,-1), the focus is at point (2,-1), and the directrix is the line x=2.

Step 1: Find the Vertex

The vertex form of a parabola is \((y-k)^2=4p(x-h)\). The vertex is the point (h, k). From the equation \((y+1)^2=-8x\), we can see that the vertex of the parabola is at point (0,-1).

Step 2: Find the Focus

The focus of the equation is at point \((h+p,k)\). We can find p by setting 4p equal to -8 which gives p=-2. Thus, the focus is at point ((0-(-2)),-1)=(2,-1).

Step 3: Find the Directrix

The directrix of the parabola is given by the equation \(x=h-p\). Substituting h=0 and p=-2, we get x=0-(-2)=2, so the directrix is the vertical line x=2.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is a key concept in understanding its geometric shape. It's essentially the point that marks the "turning point" of the parabola, where it changes direction. For a parabola described by the equation \((y-k)^2=4p(x-h)\), the vertex is located at \((h, k)\).

In this form, the vertex form, \(h\) and \(k\) are the x and y coordinates of the vertex, respectively. Identifying the vertex is crucial because it helps inform us about the orientation and position of the parabola on a graph.

• In the given exercise, the equation \((y+1)^2 = -8x\) is written in vertex form \((y+1)^2 = 4p(x-0)\).
• Here, the vertex \((h, k)\) is visibly \((0, -1)\); thus, the vertex lies at the point (0, -1) on the Cartesian plane.

Recognizing this allows us to plot the parabola more accurately, and gives clarity on the symmetry axis, which goes through the vertex.

Focus of a Parabola

The focus of a parabola offers insightful geometric information, revealing the path of incident light or projectiles converging at this point. The focus sits inside the parabola and is a fixed point that distinguishes the shape further.

The location of the focus is determined by the distance \(p\), from the vertex. The equation \((y-k)^2=4p(x-h)\) aids in locating the focus. The focus generally lies at \((h+p, k)\) when aligned horizontally, like in the given exercise.

• For the equation \((y+1)^2 = -8x\), we need to solve for \(p\) where \(4p = -8\). This gives \(p = -2\).
• Knowing \(h = 0, k = -1\), the focus calculated to be at \((0-(-2), -1) = (2, -1)\).

Understanding the focus helps highlight the parabola's reflective properties and plays a critical role in numerous physical applications such as satellite dishes and suspension bridges.

Directrix of a Parabola

The directrix of a parabola is an essential part of defining its boundary. It is a line, specifically a vertical or horizontal line, rather than a singular point. This line aids in characterizing the overall shape, residing opposite the focus with respect to the vertex.

The equation form \((y-k)^2=4p(x-h)\) is helpful here as well, as it allows you to determine the directrix. It is located at \(x = h - p\).

• In the context of the exercise, the equation \((y+1)^2 = -8x\) presents \(h=0\) and \(p=-2\).
• Substituting these into \(x = h - p\), we find the directrix at \(x = 2\).

This linear boundary line, in contrast to the parabola's curved nature, provides essential understanding of how the parabola opens and the space it occupies on a graph.