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Chapter 14: Problem 13

In Exercises 13 through 18 , find the vertex, the focus, an equation of the axis, and an equation of the directrix of the given parabola. Draw a sketch of the graph. $$ x^{2}+6 x+4 y+8=0 $$

Short Answer

Expert verified
Vertex: (-3, 1/4); Focus: (-3, -3/4); Axis: x = -3; Directrix: y = 5/4.

Step by step solution

01

Rewrite the equation

Rewrite the given equation in the standard form of a parabola. Start with \(x^2 + 6x + 4y + 8 = 0\). Move the constant and the \(y\) term to the other side: \(x^2 + 6x = -4y - 8\).
02

Complete the square

Complete the square on the \(x\) terms: \(x^2 + 6x\). Add and subtract 9 (which is \( (6/2)^2 = 9\)) on the left side: \(x^2 + 6x + 9 - 9 = -4y - 8\). Rewrite as \((x + 3)^2 - 9 = -4y - 8\).
03

Simplify the equation

Simplify to get the standard form: \((x + 3)^2 = -4y + 1\). Hence, \((x + 3)^2 = -4(y - 1/4)\) is the standard form of a vertical parabola.
04

Identify the vertex

From the equation \((x + 3)^2 = -4(y - 1/4)\), the vertex is \((-3, 1/4)\).
05

Identify the focus

Use the formula \(4p = -4\) to find the parameter \(p\), giving \(p = -1\). The focus is at \((h, k + p)\), thus the focus is at \((-3, 1/4 - 1) = (-3, -3/4)\).
06

Equation of the axis

The axis is the line passing through the vertex and perpendicular to the directrix. Hence, the axis of symmetry is \(x = -3\).
07

Equation of the directrix

The directrix is \(y = k - p\), thus \(y = 1/4 + 1 = 5/4\).
08

Graph the parabola

Sketch the graph: Draw the vertex at \((-3, 1/4)\), the focus at \((-3, -3/4)\), the axis of symmetry at \(x = -3\), and the directrix at \(y = 5/4\). Then sketch the parabola opening downwards.

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Key Concepts

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

vertex of parabola
The vertex of a parabola is a crucial point where the curve changes direction. For the given equation \((x + 3)^2 = -4(y - 1/4)\), the vertex can be identified directly from the standard form of a vertical parabola \((x-h)^2 = 4p(y-k)\). Here, \(h\) and \(k\) denote the coordinates of the vertex. In our equation, \(h = -3\) and \(k = 1/4\), so the vertex is \((-3, 1/4)\). This vertex gives us a starting point to understand the rest of the parabola's properties.
focus of parabola
The focus of a parabola is a point from which distances to any point on the parabola and the directrix are equal. In our equation \((x + 3)^2 = -4(y - 1/4)\), we find the value of \(p\) using the relationship \(4p = -4\), giving us \(p = -1\). The focus is located at \((h, k + p)\). Therefore, substituting the vertex coordinates, the focus is at \((-3, 1/4 - 1) = (-3, -3/4)\). The focus helps determine the exact shape and orientation of the parabola.
directrix of parabola
The directrix of a parabola is a line that, together with the focus, helps define the shape and position of the parabola. For our given equation \((x + 3)^2 = -4(y - 1/4)\), the directrix can be found using the formula \(y = k - p\). Given \(k = 1/4\) and \(p = -1\), the directrix is \((y = 1/4 + 1 = 5/4)\). This line runs parallel to the axis of symmetry and helps visualize the parabola's structure when graphed.
completing the square
Completing the square is a method used to transform quadratic equations into a more manageable form for identifying key properties like the vertex. For the equation \(x^2 + 6x + 4y + 8 = 0\), we isolate the quadratic and linear terms in \(x\): \(x^2 + 6x = -4y - 8\). To complete the square, add and subtract \(9\) (since \((6/2)^2 = 9)\) to get \((x + 3)^2 - 9 = -4y - 8\). Simplify this to \((x + 3)^2 = -4(y - 1/4)\). This standard form makes it much easier to identify properties like the vertex, focus, and directrix of the parabola.

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Most popular questions from this chapter

Find an equation of the parabola having the given properties. Draw a sketch of the graph. Directrix, \(x=-2\); axis, \(y=4\); length of the latus rectum is 8 .

Find an equation of the parabola having the given properties. Focus, \((0,4) ;\) directrix, \(y=-4 .\)

Find the vertex, the focus, an equation of the axis, and an equation of the directrix of the given parabola. Draw a sketch of the graph. $$ y^{2}+6 x+10 y+19=0 $$

Find the vertex, the focus, an equation of the axis, and an equation of the directrix of the given parabola. Draw a sketch of the graph. $$ y=3 x^{2}-3 x+3 $$

Find the eccentricity, center, foci, and directrices of each of the given ellipses and draw a sketch of the graph. $$ 9 x^{2}+4 y^{2}-18 x+16 y-11=0 $$
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