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Chapter 13: Problem 15

For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ x^{2}+8 y+16=0 $$

Short Answer

Expert verified
The vertex is \((0, -2)\), focus is \((0, -4)\), and directrix is \(y = 0\).

Step by step solution

01

Write the Equation in Standard Form

The given equation is \(x^2 + 8y + 16 = 0\). First, we need to rewrite it in the standard form of a parabola. Let's isolate the \(y\) terms by moving constants to the other side: \(x^2 = -8y - 16\). Next, rearrange it to get \(y\) on one side: \(x^2 = -8(y + 2)\). Now, we have the equation in the standard form \(x^2 = 4p(y - k)\), which suggests a vertical parabola.
02

Identify the Vertex

The standard form of a vertical parabola is \(x^2 = 4p(y - k)\). From the equation \(x^2 = -8(y + 2)\), it's clear that \(k = -2\) and \(h = 0\). Thus, the vertex \((h, k)\) of the parabola is \((0, -2)\).
03

Find the Parameter \(p\) and the Focus

From the standard form \(x^2 = 4p(y - k)\), we compare with \(x^2 = -8(y + 2)\) and find \(4p = -8\), hence \(p = -2\). The focus of the parabola is located \(p\) units from the vertex along the axis of symmetry. Therefore, the focus is \((0, -2 + (-2)) = (0, -4)\).
04

Determine the Directrix

The directrix is a line perpendicular to the axis of symmetry, which in this case is the vertical line \(y = k - p\). Thus, the directrix is \(y = -2 - (-2)\), simplifying to \(y = 0\).
05

Sketch the Parabola

To graph the parabola, start by plotting the vertex at \((0, -2)\). The parabola opens downward as \(p < 0\), indicating a downward curve. Plot the focus at \((0, -4)\) and draw the transverse axis vertically. The directrix is the horizontal line \(y = 0\). The vertex is equidistant between the directrix and the focus.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex
In the world of parabolas, the vertex holds a special place. It's the point where the parabola changes direction. Imagine standing at the top of a hill or the bottom of a valley. In this exercise, our parabola's equation is transformed into the standard form: \(x^2 = 4p(y - k)\). For this equation, the vertex is given as \((h, k)\). From the solution, we found \(k = -2\) and \(h = 0\), making the vertex \((0, -2)\).
This point represents the turning point on the graph. Every parabola has just one vertex, acting as a symmetrical center. Here鈥檚 a tip: for vertical parabolas like this one, the vertex is easily spotted in its equation form \((x-h)^2 = 4p(y-k)\).
  • The vertex is essential for graphing as it provides a "base".
  • It splits the parabola into two mirrored halves.
Focus
The focus of a parabola might sound like a small point but it plays a huge role in shaping the curve. Visually, it's the point inside a parabola that every point on the curve "focuses" or "aims" toward. By comparing the standard form \(x^2 = 4p(y-k)\) with our equation, we derived \(p = -2\). This means the focus is 2 units down from the vertex.
Hence the focus is at \((0, -4)\) for this parabola. Imagine a flashlight beam; the glow center (focus) shapes the direction and spread of light (parabola). The focus helps determine the parabola's "steepness" or "width".
  • The focus is always \(p\) units away from the vertex on the axis of symmetry.
  • For a vertical parabola like ours, the focus dictates where the beam hits straight down.
Directrix
Think of the directrix as a guide rail setting boundaries for a parabola. It might not touch the curve, but it's vital in its defining characteristic. The directrix is a line, perpendicular to the axis of symmetry, that lies \(p\) units from the vertex鈥攋ust opposite the focus. For our equation, we determined the directrix is \(y = 0\).
In simpler terms, it's like drawing a line under the base of a bowl, with the sides bending above it following the line鈥檚 distance. The parabola bends towards the focus but stays equidistant from this imaginary line.
  • The vertex is exactly halfway between the focus and the directrix.
  • The directrix is crucial for understanding how the parabola "scoops" or "cups."

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Most popular questions from this chapter

The point \((-1,4)\) is the midpoint of a chord of a circle whose equation is \(x^{2}+y^{2}+8 x+4 y-30=0\). Find the equation of the chord.

Vertex \((6,-4)\), symmetric with respect to the line \(y=-4\), and contains the point \((8,-3)\) $$ 2 y^{2}+16 y-x+38=0 $$

Vertex \((-9,1)\), symmetric with respect to the line \(x=-9\), and contains the point \((-8,0)\) $$ x^{2}+18 x+y+80=0 $$

For Problems \(31-50\), find an equation of the parabola that satisfies the given conditions. $$ \text { Focus }(-4,5) \text {, directrix } x=0 \quad y^{2}-10 y+8 x+41=0 $$

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ 4 x^{2}+16 x+y^{2}+2 y+1=0 $$
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