Problem 24 - Algebra for College Students

Question

For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ x^{2}+4 x-8 y-4=0 $$

Step-by-Step Solution

Answer

Vertex: (-2, -1), Focus: (-2, 1), Directrix: y = -3.

Step 1: Rewrite the Equation in Standard Form

To find the vertex, focus, and directrix, we first rewrite the given equation of the parabola in its standard form. The provided equation is \[ x^2 + 4x - 8y - 4 = 0 \]We start by isolating the terms that involve \(x\):\[ x^2 + 4x = 8y + 4 \]

Step 2: Complete the Square on x-terms

We now complete the square for the equation involving \(x\). Add and subtract the square of half of the coefficient of \(x\), which is \( (4/2)^2 = 4 \) to the left side of the equation:\[x^2 + 4x + 4 = 8y + 4 + 4\]This simplifies to:\[(x + 2)^2 = 8y + 8\]

Step 3: Simplify to Identify the Parabola Form

Further simplify the equation by setting \[-8(y + 1) = -(x + 2)^2\]So the equation becomes:\[(x + 2)^2 = 8(y + 1)\]This is the standard form of a parabola that opens upwards, which is \((x - h)^2 = 4p(y - k)\). Comparing, we identify \(h = -2\), \(k = -1\), and \(4p = 8\), so \(p = 2\).

Step 4: Find the Vertex

The vertex of the parabola is given by \((h, k)\), thus the vertex is:\[ (-2, -1) \]

Step 5: Determine the Focus

The focus is located \(p\) units from the vertex in the direction the parabola opens. Since \(p = 2\) and the parabola opens upwards, the focus is:\[ (-2, -1 + 2) = (-2, 1) \]

Step 6: Calculate the Directrix

The directrix is a line \(p\) units from the vertex in the opposite direction of the focus. Thus, the directrix is:\[ y = -1 - 2 = -3 \]

Step 7: Sketch the Parabola

Sketch the graph with vertex at \((-2, -1)\), focus at \((-2, 1)\), and the directrix at \(y = -3\). The parabola opens upwards, away from the directrix.

Key Concepts

Parabola

A parabola is a symmetrical open curve in a plane. Imagine it as the path of a ball thrown in the air. It curves gently upwards or downwards. Parabolas are an essential element of conic sections, which include circles, ellipses, and hyperbolas. They are special because of their unique properties.

• Parabolas have a single axis of symmetry, meaning they fold perfectly in half along a line.
• Their shape is determined by a quadratic equation in the form of \( (x - h)^2 = 4p(y - k) \) for a vertical parabola or \( (y - k)^2 = 4p(x - h) \) for a horizontal one.

Understanding the structure of this equation helps us find important features like the vertex, focus, and directrix. Parabolas are seen frequently in real life, such as the design of satellite dishes and the paths rockets take.

Vertex

The vertex is the turning point of the parabola. In simpler terms, it is the highest or lowest point, depending on whether the parabola opens upwards or downwards.
To find the vertex from the equation \((x-h)^2 = 4p(y-k)\), we identify \(h\) and \(k\) as the coordinates of the vertex. It deeply signifies the balance point of the parabola.

• For our problem, the equation was \((x + 2)^2 = 8(y + 1)\).
• By comparing with the standard form, the vertex is at \((-2, -1)\).

Knowing the vertex helps us in sketching the parabola’s graph. It is also crucial because it serves as a reference point for finding the other important features like focus and directrix.

Focus

The focus of a parabola is a single point from which distances are measured in relation to the directrix. This might sound tricky, but just know that the focus helps in defining how the parabola curves around.

• The focus lies inside the parabola, with the parabola wrapping around it.
• It's exactly \(p\) units from the vertex, where \(p\) comes from the equation \(4p = 8\), thus \(p = 2\).
• For our specific parabola, the focus is at \((-2, 1)\), which is achieved by moving \(2\) units up from the vertex \((-2, -1)\).

The focus along with the directrix determines the parabola's steepness and orientation. It’s like a hidden heart of the curve, directing how the parabola will open.

Directrix

The directrix is a straight line that works together with the focus to maintain the shape of the parabola. While the focus lies inside the curve, the directrix keeps an equal measurement but outside.
The directrix is useful in defining the parabola's reflection property, where any point on the parabola is equidistant to both the focus and the directrix.

• In our example, since the focus is \(2\) units above the vertex, the directrix is \(2\) units below.
• This leads us to the directrix being at \(y = -3\).

Understanding the directrix gives a comprehensive view of the parabola's symmetry. It acts like an outer boundary to the soaring shape of the parabola, balancing its stretch opposite to the focus.