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Chapter 10: Problem 14

Find the vertex, focus, and directrix of the parabola, and sketch its graph. $$ y^{2}+6 y+8 x+25=0 $$

Short Answer

Expert verified
The vertex of the parabola is at (0, -3), the focus is at (2, -3) and the directrix is the line x = 2.

Step by step solution

01

Rearrange the equation

First, rewrite the equation so we can have all the y terms on one side and x term on the other. This gives us: \(y^2 + 6y = -8x - 25 \)
02

Complete the square for y

The formula (y - k)^2 gives us a trinomial that includes the term -2ky and the square k^2 so we need to complete the square by adding a term to make our left side into a perfect square: \(y^2 + 6y + (6/2)^2 = (y+3)^2 \). To keep the equality, the same term must be added to the right side. So we get: \((y+3)^2 = -8x - 25 + (3)^2 \) or \((y+3)^2 = -8x + 16 \)
03

Identify the parabola variables

This gives us a parabolic equation in the form: \((y - k)^2 = 4a(x - h)\). Comparing the two, we identify k = -3, h = 2 (as there is no x - h form on our right side, we should understand it as x - 0 which corresponds to h = 0) and 4a = -8 (which can be understood from the coefficient of our x term). This allows us to determine a = -2.
04

Determine the vertex, focus, and directrix

From above, the vertex becomes (h,k) = (0,-3). The focus is at (h+a, k) = (0-(-2), -3) = (2,-3) and the directrix is the line x = h - a = 0 - -2 = 2.

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Key Concepts

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

Vertex
The vertex of a parabola is a crucial point that indicates where the curve changes direction. It's the point where the parabola is either at its highest or lowest point, depending on its orientation (upward or downward for vertical parabolas, and leftward or rightward for horizontal ones).
In the context of our exercise, the vertex can be represented as \((h, k)\) according to the transformed parabolic equation format \((y - k)^2 = 4a(x - h)\).
This means that, after completing the square, we've identified \(h = 0\) and \(k = -3\). Therefore, the vertex of the parabola \(y^2 + 6y + 8x + 25 = 0\) is at \(0, -3\).
  • Vertex determines where the parabola changes course.
  • Located at the point \(h, k\).
  • In this example: vertex is \(0, -3\).
Focus
The focus of a parabola is a significant component that helps define its shape. The focus is a fixed point that, along with the directrix, describes how far any point on the parabola is from the focus compared to the directrix.
It's placed at a distance \(a\) from the vertex either inside or outside the parabola, depending on whether it's facing left, right, up, or down. For a parabola defined by \((y - k)^2 = 4a(x - h)\), the focus is at the point \( (h + a, k) \).
Therefore, in this problem, where \(a = -2\), the focus is at the point \(2, -3\).
  • Defines the shape and direction of a parabola.
  • Located at a distance from the vertex.
  • In this example: focus is at \(2, -3\).
Directrix
The directrix is a line that, together with the focus, defines a parabola. The constant distance from any point on the parabola to the focus and the directrix gives the parabola its distinct shape.
For a parabola with the equation \((y - k)^2 = 4a(x - h)\), the directrix is a vertical line determined by the equation \(x = h - a\).
This means, using the values from the example, the directrix of our parabola is the line \(x = 2\).
  • Directrix is a perpendicular line to the axis of symmetry.
  • Aids in defining the parabola's curve.
  • For this exercise: directrix is \(x = 2\).
Completing the Square
Completing the square is a mathematical technique used to transform a quadratic equation into a form that easily identifies a parabola's vertex. The method involves creating a perfect square trinomial from a quadratic expression.
We started with \(y^2 + 6y\) in its standard form and needed to express it as \((y+3)^2\) to match the standard parabola vertex form \((y - k)^2 = 4a(x - h)\).
By completing this square and adding \((3)^2\) to both sides, we altered the original equation to \((y+3)^2 = -8x + 16\), a transformation that lets us easily find the vertex and other key properties of the parabola.
  • Transforms quadratics to reveal parabolic form.
  • Enables easy identification of vertex.
  • Applied technique: transmuted \(y^2 + 6y\) into \((y+3)^2\).

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Most popular questions from this chapter

Use the result of Exercise 108 to find the angle \(\psi\) between the radial and tangent lines to the graph for the indicated value of \(\theta\). Use a graphing utility to graph the polar equation, the radial line, and the tangent line for the indicated value of \(\theta\). Identify the angle \(\psi\). \(\begin{array}{ll} \text { Polar Equation } & \text { Value of } \theta \end{array}\) $$ r=2 \cos 3 \theta \quad \theta=\pi / 4 $$

The planets travel in elliptical orbits with the sun as a focus, as shown in the figure. (a) Show that the polar equation of the orbit is given by \(r=\frac{\left(1-e^{2}\right) a}{1-e \cos \theta}\) where \(e\) is the eccentricity. (b) Show that the minimum distance (perihelion) from the sun to the planet is \(r=a(1-e)\) and the maximum distance \((\) aphelion \()\) is \(r=a(1+e)\).

Show that the polar equation for \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) is \(r^{2}=\frac{-b^{2}}{1-e^{2} \cos ^{2} \theta} \cdot\)

Sketch a graph of the polar equation. $$ r \equiv 2 $$

Consider a projectile launched at a height \(h\) feet above the ground and at an angle \(\theta\) with the horizontal. If the initial velocity is \(v_{0}\) feet per second, the path of the projectile is modeled by the parametric equations \(x=\left(v_{0} \cos \theta\right) t\) and \(y=h+\left(v_{0} \sin \theta\right) t-16 t^{2}\). The center field fence in a ballpark is 10 feet high and 400 feet from home plate. The ball is hit 3 feet above the ground. It leaves the bat at an angle of \(\theta\) degrees with the horizontal at a speed of 100 miles per hour (see figure). (a) Write a set of parametric equations for the path of the ball. (b) Use a graphing utility to graph the path of the ball when \(\theta=15^{\circ} .\) Is the hit a home run? (c) Use a graphing utility to graph the path of the ball when \(\theta=23^{\circ} .\) Is the hit a home run? (d) Find the minimum angle at which the ball must leave the bat in order for the hit to be a home run.
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