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Chapter 11: Problem 14

Graph the parabolas. In each case, specify the focus, the directrix, and the focal width. Also specify the vertex. $$y^{2}+2 y+8 x+17=0$$

Short Answer

Expert verified
Vertex: (-2, -1), Focus: (-4, -1), Directrix: x = 0, Focal Width: 8.

Step by step solution

01

Transform the Equation

Start by rearranging the given equation into a standard form of a parabola. Given: \[ y^2 + 2y + 8x + 17 = 0 \]First, isolate the terms involving \(x\): \[ y^2 + 2y = -8x - 17 \] Now, complete the square for the \(y\)-terms.
02

Completing the Square

To complete the square, take the coefficient of \(y\), which is 2, divide it by 2 and square it. Add and subtract this result inside the equation:\[ y^2 + 2y + \left(\frac{2}{2}\right)^2 - \left(\frac{2}{2}\right)^2 = -8x - 17 \]This simplifies to:\[ (y+1)^2 - 1 = -8x - 17 \]Then:\[ (y+1)^2 = -8x - 16 \]
03

Converting to Standard Form of a Parabola

Now the equation resembles the standard form of a parabola opening to the left: \[ (y-k)^2 = 4p(x-h) \]Comparing \((y+1)^2 = -8(x+2)\), we get:- \( (y+1)^2 \rightarrow (y - (-1))^2 \) thus \(k = -1\)- \(-8(x+2) \rightarrow 4p = -8\) thus \(h = -2\) and \(p = -2\)
04

Identify Vertex, Focus, and Directrix

From the standard form equation, identify:- **Vertex**: \((h, k) = (-2, -1)\)- **Focus**: Since the parabola opens to the left, focus is \((h + p, k) = (-2 - 2, -1) = (-4, -1)\)- **Directrix**: Vertical line given by \(x = -2 - (-2) = 0\)
05

Determine Focal Width

The focal width of a parabola is given by the absolute value of \(4p\). From \(4p = -8\): the focal width is:\[ \text{Focal Width} = |4p| = |8| = 8 \]

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Key Concepts

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

Parabola
A parabola is a U-shaped curve that can open upwards, downwards, or sideways depending on its orientation. It is a conic section created when a plane intersects a cone at a particular angle.
The standard equation of a parabola can take different forms based on its axis of symmetry:
  • Vertical orientation (opens up or down): \( y = ax^2 + bx + c \)
  • Horizontal orientation (opens left or right): \( x = ay^2 + by + c \)
For horizontal parabolas, such as in this exercise, the vertex form equation is \( (y-k)^2 = 4p(x-h) \). This form makes it easy to identify the vertex, the focus, and the directrix. Understanding these features helps graph the parabola accurately.
Focus and Directrix
The focus and directrix are key components that define a parabola. The focus is a point located inside the parabola, which equidistant from every point on the parabola to its corresponding point on the directrix.
Let's break this down:
  • **Focus**: Given a parabola in the form \( (y-k)^2 = 4p(x-h) \), the focus for a horizontal parabola is located at \( (h + p, k) \). In the example exercise, the focus is \( (-4, -1) \), meaning the parabola opens to the left.
  • **Directrix**: This is a line that lies outside the parabola. For our horizontal parabola, the directrix is a vertical line defined by \( x = h - p \). In the exercise, the directrix is \( x = 0 \).
These components work together to provide a complete picture of how the parabola is structured and oriented in the coordinate plane.
Vertex Form
The vertex form of a parabola is a way to express the equation that makes it straightforward to identify its vertex. In this specific problem, the vertex form emerges as \( (y+1)^2 = -8(x+2) \).
  • The h and k in the vertex form \( (y-k)^2 = 4p(x-h) \) represent the coordinates of the vertex: \( (h, k) \).
  • For the exercise, the vertex is at \( (-2, -1) \).
Recognizing and using the vertex form simplifies the process of pinpointing the precise location of the parabola's vertex. This form is crucial when graphing since it directly tells you where the "tip" of the parabola is positioned.
Focal Width
The focal width of a parabola pertains to the width of the parabola at the level of the focus. It gives an idea of how "wide" the parabola is near the focus and vertex.
  • The focal width is calculated as the absolute value of \( 4p \), where \( p \) is the distance between the vertex and the focus.
  • In the exercise, \( 4p = -8 \), thus the focal width is \( |8| = 8 \).
This number is always positive, representing a real distance. By understanding the focal width, one can better understand the shape and spread of the parabola in relation to its focus.

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Most popular questions from this chapter

(a) Verify that the point \(P(6,4 \sqrt{3})\) lies on the hyperbola \(16 x^{2}-9 y^{2}=144\) (b) In Example \(1,\) we found that the foci of this hyperbola were \(F_{1}(-5,0)\) and \(F_{2}(5,0) .\) Compute the lengths \(F_{1} P\) and \(F_{2} P\), where \(P\) is the point \((6,4 \sqrt{3})\) (c) Verify that \(\left|F_{1} P-F_{2} P\right|=2 a\)

Find the equation of the line that is tangent to the hyperbola at the given point. Write your answer in the form \(y=m x+b\). $$3 x^{2}-y^{2}=12 ;(4,6)$$

This exercise outlines the steps required to show that the equation of the tangent to the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) at the point \(\left(x_{1}, y_{1}\right)\) on the ellipse is \(\left(x_{1} x / a^{2}\right)+\left(y_{1} y / b^{2}\right)=1\) (a) Show that the equation \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is equivalent to $$b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$$(1) Conclude that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse if and only if $$b^{2} x_{1}^{2}+a^{2} y_{1}^{2}=a^{2} b^{2}$$(2) (b) Subtract equation ( \(2)\) from equation (1) to show that $$b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0$$(3) Equation ( 3 ) is equivalent to equation ( 1 ) provided only that \(\left(x_{1}, y_{1}\right)\) lies on the ellipse. In the following steps, we will find the algebra much simpler if we use equation ( 3 ) to represent the ellipse, rather than the equivalent and perhaps more familiar equation (1). (c) Let the equation of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) be $$y-y_{1}=m\left(x-x_{1}\right)$$ Explain why the following system of equations must have exactly one solution, namely, \(\left(x_{1}, y_{1}\right);\) $$\left\\{\begin{array}{ll}b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2}\left(y^{2}-y_{1}^{2}\right)=0 & (4) \\ y-y_{1}=m\left(x-x_{1}\right) & (5)\end{array}\right.$$ (d) Solve equation ( 5 ) for \(y\), and then substitute for \(y\) in equation (4) to obtain \(b^{2}\left(x^{2}-x_{1}^{2}\right)+a^{2} m^{2}\left(x-x_{1}\right)^{2}+2 a^{2} m y_{1}\left(x-x_{1}\right)=0\) (e) Show that equation (6) can be written \(\left(x-x_{1}\right)\left[b^{2}\left(x+x_{1}\right)+a^{2} m^{2}\left(x-x_{1}\right)+2 a^{2} m y_{1}\right]=0\) (f) Equation (7) is a quadratic equation in \(x,\) but as was pointed out earlier, \(x=x_{1}\) must be the only solution. (That is, \(x=x_{1}\) is a double root.) Thus the factor in brackets must equal zero when \(x\) is replaced by \(x_{1}\) Use this observation to show that $$m=-\frac{b^{2} x_{1}}{a^{2} y_{1}}$$ This represents the slope of the line tangent to the ellipse at \(\left(x_{1}, y_{1}\right)\) (g) Using this value for \(m,\) show that equation (5) becomes $$b^{2} x_{1} x+a^{2} y_{1} y=b^{2} x_{1}^{2}+a^{2} y_{1}^{2}$$(8) (h) Now use equation (2) to show that equation (8) can be written $$\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1$$ which is what we set out to show.

If \(\overline{P Q}\) is a focal chord of the parabola \(x^{2}=4 p y\) and the coordinates of \(P\) are \(\left(x_{0}, y_{0}\right),\) show that the coordinates of \(Q\) are $$\left(\frac{-4 p^{2}}{x_{0}}, \frac{p^{2}}{y_{0}}\right)$$

(a) Verify that the points \(A(5,1), B(4,-2),\) and \(C(-1,3)\) all lie on the ellipse \(x^{2}+3 y^{2}=28\). (b) Find a point \(D\) on the ellipse such that \(\overline{C D}\) is parallel to \(\overline{A B}\). (c) If \(O\) denotes the center of the ellipse, show that the triangles \(O A C\) and \(O B D\) have equal areas. Suggestion: In computing the areas, the formula given at the end of Exercise 34 in Section 1.4 is useful.
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