Chapter 14: Problem 15
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. $$ y^{2}+6 x+10 y+19=0 $$
Short Answer
Expert verified
The vertex is (1, -5), the focus is \(\left(-\frac{1}{2}, -5\right)\), the axis is \( y = -5 \), and the directrix is \( x = \frac{5}{2} \).
Step by step solution
01
- Rearrange the equation into standard form
Start by rearranging and completing the square to convert the given equation into a standard form. Given equation is \( y^{2} + 6x + 10y + 19 = 0 \) First, rewrite the equation by isolating the terms involving y on one side: \( y^{2} + 10y = -6x - 19 \)
02
- Complete the square
To complete the square for the y terms, take half of the coefficient of y (which is 10), square it, and add and subtract that value on the left-hand side: \( y^{2} + 10y + 25 - 25 = -6x - 19 \) Thus, \( (y + 5)^{2} = -6x - 19 + 25 \) which simplifies to \( (y + 5)^{2} = -6x + 6 \) Therefore, \( (y + 5)^{2} = -6(x - 1) \)
03
- Identify the standard form
The equation \( (y + 5)^{2} = -6(x - 1) \) is in the form \( (y - k)^{2} = 4p(x - h) \) Here, comparing, we get: \( h = 1 \), \( k = -5 \), and \( 4p = -6 \) Solving this, \( p = -\frac{3}{2} \)
04
- Find the vertex
The vertex \( (h, k) \) from the standard form is \( (1, -5) \)
05
- Calculate the focus
For a parabola in the form \( (y - k)^{2} = 4p(x - h) \), the focus is located at \((h + p, k)\). Substitute values, focus location is \( 1 - \frac{3}{2}, -5 \), which simplifies to \(-\frac{1}{2}, -5 \)
06
- Find the equation of the axis
The axis of the parabola is the vertical line through the vertex, so the equation of the axis is \( y = -5 \)
07
- Find the equation of the directrix
For a parabola \( (y - k)^{2} = 4p(x - h) \), the directrix is a vertical line at \( x = h - p \). Using the values, the directrix is located at \( x = 1 + \frac{3}{2} \), which simplifies to \( x = \frac{5}{2} \)
08
- Draw the sketch
Sketch the graph indicating the vertex at \((1, -5)\), the focus at \(\left(-\frac{1}{2}, -5\right)\), the axis at \( y = -5 \) and the directrix at \( x = \frac{5}{2} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
completing the square
Completing the square is a method used to simplify quadratic equations by turning part of the equation into a perfect square trinomial. This method is useful when transforming the equation of a parabola into its standard form. To understand completing the square, follow these simple steps:
First, isolate the quadratic and linear terms. For example, from the equation given: \[ y^{2} + 6x + 10y + 19 = 0 \]
Separate terms with y:
\[ y^{2} + 10y = -6x - 19 \]
Next, take half of the coefficient of y, square it, and add/subtract it back to the equation:
\[ \text{Coefficient of } y = 10, \text{Half of } 10 = 5, 5^{2} = 25 \]
So, add and subtract 25:
\[ y^{2} + 10y + 25 - 25 = -6x - 19 \]
This converts perfectly to:
\[ (y + 5)^{2} - 25 = -6x - 19 \]
Finally, rearrange the terms to form:
\[ (y + 5)^{2} = -6(x - 1) \]
Completing the square helps us identify key features of the parabola, such as its vertex and focus.
First, isolate the quadratic and linear terms. For example, from the equation given: \[ y^{2} + 6x + 10y + 19 = 0 \]
Separate terms with y:
\[ y^{2} + 10y = -6x - 19 \]
Next, take half of the coefficient of y, square it, and add/subtract it back to the equation:
\[ \text{Coefficient of } y = 10, \text{Half of } 10 = 5, 5^{2} = 25 \]
So, add and subtract 25:
\[ y^{2} + 10y + 25 - 25 = -6x - 19 \]
This converts perfectly to:
\[ (y + 5)^{2} - 25 = -6x - 19 \]
Finally, rearrange the terms to form:
\[ (y + 5)^{2} = -6(x - 1) \]
Completing the square helps us identify key features of the parabola, such as its vertex and focus.
standard form of a parabola
The standard form of a parabola offers a clear view of key characteristics like the vertex and the direction it opens. For a vertical parabola, the standard form is:
\[ (y - k)^{2} = 4p(x - h) \]
In this equation:
From our given parabola:
\[ (y + 5)^{2} = -6(x - 1) \]
By comparing, we see:
From this, solve for p:
\[ 4p = -6 \]\[ p = -\frac{3}{2} \]
So the standard form tells us the vertex is at (1, -5) and the parabola opens to the left because p is negative.
\[ (y - k)^{2} = 4p(x - h) \]
In this equation:
- (h, k) is the vertex of the parabola.
- p is the distance from the vertex to the focus.
From our given parabola:
\[ (y + 5)^{2} = -6(x - 1) \]
By comparing, we see:
- h = 1
- k = -5
- 4p = -6
From this, solve for p:
\[ 4p = -6 \]\[ p = -\frac{3}{2} \]
So the standard form tells us the vertex is at (1, -5) and the parabola opens to the left because p is negative.
focus and directrix of a parabola
The focus and directrix of a parabola give us insights into its geometric properties. The focus is a specific point that lies inside the parabola, serving as a point from which distances are measured that define the curve. The directrix is a line outside the parabola used in the same way.
For the given standard form:
\[ (y + 5)^{2} = -6(x - 1) \]
The focus point is:
\[ (h + p, k) = \left(1 - \frac{3}{2}, -5\right) = \left(-\frac{1}{2}, -5\right) \]
The directrix line is:
\[ x = h - p = 1 + \frac{3}{2} = \frac{5}{2} \]
Understanding the focus and directrix helps sketch the parabola and visualize its shape and direction.
- For the form \( (y - k)^{2} = 4p(x - h) \)
- Focus: Located at \( (h + p, k) \)
- Directrix: Line given by \( x = h - p \)
For the given standard form:
\[ (y + 5)^{2} = -6(x - 1) \]
The focus point is:
\[ (h + p, k) = \left(1 - \frac{3}{2}, -5\right) = \left(-\frac{1}{2}, -5\right) \]
The directrix line is:
\[ x = h - p = 1 + \frac{3}{2} = \frac{5}{2} \]
Understanding the focus and directrix helps sketch the parabola and visualize its shape and direction.
parabola axis
The parabola axis, also known as the axis of symmetry, is a line that divides the parabola into two mirror-image halves. This axis always goes through the vertex and is perpendicular to the direction in which the parabola opens.
Given the standard form of our parabola:
\[ (y + 5)^{2} = -6(x - 1) \]
The axis of symmetry passes through the vertex at (1, -5). Since our parabola opens horizontally, the axis of symmetry is a horizontal line. Thus, the equation of the axis is:
\[ y = -5 \]
The axis helps us to accurately draw and understand the parabola’s orientation and properties.
- For vertical parabolas: The axis is a vertical line.
- For horizontal parabolas: The axis is a horizontal line.
Given the standard form of our parabola:
\[ (y + 5)^{2} = -6(x - 1) \]
The axis of symmetry passes through the vertex at (1, -5). Since our parabola opens horizontally, the axis of symmetry is a horizontal line. Thus, the equation of the axis is:
\[ y = -5 \]
The axis helps us to accurately draw and understand the parabola’s orientation and properties.