Problem 40 - Thomas Calculus

Question

The parabola \(x^{2}=-4 y\) is shifted left 1 unit and up 3 units to generate the parabola \((x+1)^{2}=-4(y-3)\) a. Find the new parabola's vertex, focus, and directrix. b. Plot the new vertex, focus, and directrix, and sketch in the parabola.

Step-by-Step Solution

Answer

Vertex: (-1, 3); Focus: (-1, 2); Directrix: y = 4.

Step 1: Identify the Vertex

The original equation of the parabola is given as \(x^2 = -4y\). In standard form, it is \((x-h)^2 = -4p(y-k)\) where \((h,k)\) is the vertex of the parabola. From the equation \((x+1)^2 = -4(y-3)\), we can see that \(h = -1\) and \(k = 3\), so the vertex is \((-1, 3)\).

Step 2: Find the Parameter \(p\)

From the standard form equation \((x+1)^2 = -4(y-3)\), compare it with \((x-h)^2 = -4p(y-k)\). Notice that \(-4 = -4p\), so \(p = 1\).

Step 3: Determine the Focus

The focus of a parabola \((x-h)^2 = -4p(y-k)\) is given by \((h, k-p)\). Substitute \(h = -1\), \(k = 3\), and \(p = 1\) into this form. Thus, the focus is at \((-1, 3-1) = (-1, 2)\).

Step 4: Compute the Directrix

The directrix of the parabola \((x-h)^2 = -4p(y-k)\) is given by the equation \(y = k+p\). Substitute \(k = 3\) and \(p = 1\), so the directrix is \(y = 3 + 1 = 4\).

Step 5: Plot the Points and Sketch the Parabola

Plot the vertex \((-1, 3)\), the focus \((-1, 2)\), and the line \(y = 4\) as the directrix on a graph. Sketch the parabola opening downwards, because the coefficient of \(y\) in the original \(-4(y-3)\) is negative.

Key Concepts

Vertex

In the realm of parabolas, the vertex holds a place of great importance. It is the point where the parabola turns or "bends." In mathematical terms, the vertex serves as the coordinate where the parabola reaches its minimum or maximum value. For our parabola, evaluated from the equation yielding the form equation \((x+1)^2 = -4(y-3)\), this can be observed clearly. Based on the standard form, \((x-h)^2 = -4p(y-k)\), we find our vertex at \((-1, 3)\). This tells us two main things:

• The parabola is shifted 1 unit to the left, as indicated by \(h = -1\).
• It is also shifted 3 units up, given that \(k = 3\).

Understanding the vertex is crucial, as it helps to determine the symmetry and the directional features of the parabola on the coordinate plane.

Focus

The focus of a parabola is another key concept that adds depth to its representation. Geometrically, the focus is a special point located inside the parabola. It is significant because it helps in defining the shape and direction of the parabola. Mathematically, for the given equation of our parabola, \((x+1)^2 = -4(y-3)\), we can locate the focus by the formula \((h, k-p)\). Here,

• The vertex \((h, k)\) is \((-1, 3)\),
• and \(p = 1\) was found from comparing coefficients.

Thus, substituting in these values: \[ (-1, 3-1) = (-1, 2) \]. This informs us that the parabola leans toward its focus as it "opens," and this specific point helps in drawing the parabolic shape accurately. The focus is always positioned inside the curve and plays a critical role in its applications, including reflectivity properties in physics and engineering.

Directrix

Along with the vertex and focus, the directrix completes our map of the parabola. It is a straight line that acts as a reference for the parabola's curve but lies outside of it. The placement of the directrix is parallel to the axis of symmetry of the parabola, and it is perpendicular to the direction in which the parabola opens. For our equation \((x+1)^2 = -4(y-3)\),the directrix can be determined using the formula \(y = k+p\), where

• \(k\) is the vertical shift, \(k = 3\),
• and \(p = 1\) is derived from the parabola's standard form.

So, substituting these values, we have: \[ y = 3 + 1 = 4 \]. This line \(y = 4\) is the directrix for our parabola, and it tells us that the parabola opens downward below this line. The distance from any point on the parabola to the focus is equal to the distance to the directrix, forming the basis of the parabola's reflective property. Understanding where the directrix is in relation to the focus and vertex is key to understanding the parabolic shape and its orientation.