Problem 40 - Thomas Calculus

Question

You may wish to review Section 1.2 before solving Exercises \(39-56\) The parabola \(x^{2}=-4 y\) is shifted left 1 unit and up 3 units to generate the parabola \((x+1)^{2}=-4(y-3)\) $$ \begin{array}{l}{\text { a. Find the new parabola's vertex, focus, and directrix. }} \\ {\text { b. Plot the new vertex, focus, and directrix, and sketch in the }} \\ {\text { parabola. }}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Identify New Vertex

The original equation of the parabola is shifted left by 1 unit and up by 3 units. The standard form of a parabola is \( (x-h)^2 = 4p(y-k) \), where \( (h, k) \) is the vertex. For the equation \( (x+1)^{2}=-4(y-3) \), the vertex is \( (-1, 3) \).

Step 2: Determine Focus and Directrix

The equation \( (x+1)^{2} = -4(y-3) \) indicates that \( 4p = -4 \), so \( p = -1 \). Since the parabola opens downward, the focus is located \( p \) units down from the vertex. Therefore, the focus is \( (-1, 3-1) = (-1, 2) \). The directrix is \( p \) units above the vertex, so the equation for the directrix is \( y = 3 + 1 = 4 \).

Step 3: Sketch the Parabola

Plot the vertex at \( (-1, 3) \) and the focus at \( (-1, 2) \). Draw the line \( y = 4 \) as the directrix. Since the focus is below the vertex, draw the parabola opening downward, ensuring it curves towards the focus and away from the directrix.

Key Concepts

Vertex

The vertex of a parabola is a significant point because it represents the highest or lowest point on the graph, depending on the direction the parabola opens. For the given parabola equation \[ (x+1)^2 = -4(y-3) \]the vertex is \[ (-1, 3) \].Here's why:

• A parabola in vertex form, \[ (x-h)^2 = 4p(y-k) \], has its vertex at the point \((h, k)\).
• Here, the vertex is the point where the parabola changes direction.
• For our equation, \((h, k) = (-1, 3)\), indicating the parabola is shifted 1 unit left and 3 units up from the origin.

The vertex gives us a precise way to know the turning point of the parabola and assists in sketching its graph.Understanding the position of the vertex allows us to easily determine the basic shape and direction of the parabola.

Focus

The focus of a parabola is a point from which the distances to any point on the parabola are perfect, almost like a spotlight that guides the curve of the shape. For the parabola equation \[(x+1)^2 = -4(y-3)\],the focus is at \[(-1, 2)\].Here’s how we find it:

• If \[4p = -4\], then \[p = -1\], which tells us how far the focus is from the vertex.
• The negative value of \[p\] shows that the parabola opens downward, meaning the focus is below the vertex.
• Thus, from the vertex \[(-1, 3)\], moving 1 unit down gives us the focus at \[(-1, 2)\].

The focus is crucial because every point on the parabola is equidistant from both the focus and the directrix. This constant distance property helps us maintain the shape of the parabola.

Directrix

The directrix of a parabola is a line that, together with the focus, helps define the curve. For the parabola equation\[(x+1)^2 = -4(y-3)\],the directrix is \[y = 4\].Let’s explain:

• The directrix is always outside the path of the parabola. It is a straight line opposite to the focus in terms of distance from the vertex.
• Since \[p = -1\], this means our directrix is 1 unit above the vertex, hence \[y = 4\].
• This line is significant because it lies parallel to the x-axis when the parabola opens up or down.

In essence, the directrix, combined with the focus, provides a balanced framework for the parabolic shape, ensuring that any point on the curve is the same distance from both the focus and this line.