Problem 39 - Precalculus with Limits

Question

In Exercises 33-46, find the vertex, focus, and directrix of the parabola, and sketch its graph. \((x-1)^2 + 8(y+2) = 0\)

Step-by-Step Solution

Answer

Vertex: (1,-2), Focus: (1,-4), Directrix: y = 0

Step 1: Determine the Vertex

The vertex of the parabola is (h, k). Identify the values of h and k by comparing the given equation \( (x-1)^2 = -8(y+2) \) with the standard form. Here, the vertex is (1, -2).

Step 2: Identify the value of a

The value of 4a is the coefficient of the \( y \) term. From the equation \( (x-1)^2 = -8(y+2) \), we have 4a = -8. Solving for \( a \) gives us \( a = -2 \).

Step 3: Find the Focus

The focus lies a units along the parabola's axis of symmetry. Here, since a is negative, we move 'a' units down from the vertex. So, the coordinates of the focus are (1, -2 - (-2)) = (1, -4).

Step 4: Find directrix

The directrix is a line that is 'a' units above the vertex (since a is negative here). Therefore, \( y = -2 + 2 = 0 \) is the equation of the directrix.

Step 5: Sketch the Graph

Plot the vertex, focus, and directrix on a graph. The parabola opens downwards. Use these points and the downwards opening to sketch the graph.

Key Concepts

Vertex of Parabola

Think of the vertex of a parabola as the turning point in a thrilling story—it's the peak or pit, where things change direction. When you're looking at a parabola on a graph, the vertex is that poignant spot where the curve stops going down and starts climbing up, or vice versa. It marks the minimum or maximum of the parabola's arc.

In algebraic terms, for a quadratic equation written in the standard form \( y = ax^2 + bx + c \) or \( (x - h)^2 = 4a(y - k) \), the vertex is presented as the point \( (h, k) \). If your equation is tossed around a bit, don't fret—just shuffle it back into one of these standard forms to spot the vertex. It's a bit like solving a riddle: find the proper form and voilà, the vertex appears. In our exercise, we shuffled \( (x-1)^2 = -8(y+2) \) to reveal the vertex (1, -2), a crucial first step in understanding our parabola.

Focus of Parabola

Dive into the heart of a parabola, and you'll find the focus—a point that might seem mystical but is quite the practical character. The focus isn't just floating around; it's always strategically positioned. If you were to draw lines from every point on the parabola to this focus, you'd notice they all bounce off the curve at equal angles. It's like socializing at a party; from the focus, every point on the parabola gets equal attention.

Now, finding the focus involves understanding a fundamental characteristic: the distance from the vertex to the focus is measured by the value \( |a| \) in the equation \( (x - h)^2 = 4a(y - k) \). If \( a > 0 \), your parabola opens up and the focus is above the vertex. If \( a < 0 \), like in our equation with \( a = -2 \), it's opening down and the focus is below the vertex. In the exercise, we found the focus at (1, -4) by moving \( a \) units from the vertex.

Directrix of Parabola

While the focus lies inside the parabola, the directrix line is like a boundary line lying outside, never to be crossed by the curve. It's essential because every point on the parabola keeps an equal distance from it as from the focus—like the perfect balance in a tug of war.

For an upright or an inverted parabola, the directrix will be a horizontal line, and its equation is simple: \( y = k - a \) when the parabola opens upwards, and \( y = k + a \) when it opens downwards. This 'a' value is the same as in the parabola's equation, and it's always about keeping the perfect balance with the focus. Our example had a directrix at \( y = 0 \), which we found by adding 'a' to the vertex's y-coordinate because our parabola opens downwards and we want to maintain that balance.

Graphing Parabolas

Sketching a parabola is a bit like drawing a portrait. You start with key features—the vertex, focus, and directrix—and then connect the dots smoothly into a U-shaped curve. But remember, a parabola can also flip its 'U' upside down! Whether it smiles at you with a U or frowns with an inverted one depends on its mood, or more accurately, the sign of 'a'.

Graphing starts with plotting the vertex as a central point—like we did in the exercise at (1, -2)—then adding the focus, directrix, and a few more points by plugging in x-values and solving for y. Connect these dots with care, giving your curve that classic parabolic shape. In our case, we created a downward opening parabola, its vertex frowning on the graph, a portrait of symmetrical beauty.

Standard Form of Parabola Equation

When you face a parabola, you want it to make sense right away, and the standard form is like its identity card—it tells you everything you need to know at a glance. The equation \( (x - h)^2 = 4a(y - k) \) might look like a secret code, but it's actually a decoder for the parabola's characteristics.

The standard form neatly packages the vertex into \( (h, k) \), tells you about the opening direction and width through \( a \), and even shows the axis of symmetry. Adjusting 'a' stretches or squeezes the parabola and flips it if 'a' is negative. The beauty of this form is in its clarity—it immediately gives you the tools to find the focus, directrix, and to sketch the graph. With practice, you'll unlock any parabola equation's secrets just as we unveiled the mysteries of \( (x-1)^2 = -8(y+2) \) in our textbook problem.