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Chapter 18: Problem 66

The equation of the directrix of the parabola \(y^{2}+4 y+\) \(4 x+2=0\) is: (A) \(x=-1\) (B) \(x=1\) (C) \(x=-\frac{3}{2}\) (D) \(x=\frac{3}{2}\)

Short Answer

Expert verified
The directrix is \(x = \frac{3}{2}\) (Option D).

Step by step solution

01

Identify the form of the parabola equation

The given equation is in the form of a rotated parabola equation: \(y^2 + 4y + 4x + 2 = 0\). Our goal is to rearrange this to identify its vertex form \((y-k)^2 = 4p(x-h)\), which then helps in finding its directrix.
02

Complete the square for the y-terms

First, group and complete the square for the terms involving \(y\). We start with \(y^2 + 4y\). Completing the square yields: \((y+2)^2 - 4\). Thus, the equation becomes \((y+2)^2 - 4 + 4x + 2 = 0\).
03

Simplify and rearrange the equation

Simplify the equation: \((y+2)^2 - 4 + 4x + 2 = 0\) becomes \((y+2)^2 = -4x + 2\). Rearrange to get \((y+2)^2 = -4(x - \frac{1}{2})\).
04

Identify the components of the parabola equation

The equation \((y+2)^2 = -4(x - \frac{1}{2})\) indicates a parabola that opens to the left. Here, \(h = \frac{1}{2}\) and \(p = -1\).
05

Determine the equation of the directrix

The directrix of a parabola \((y-k)^2 = 4p(x-h)\) is given by the equation \(x = h - p\). Substituting \(h = \frac{1}{2}\) and \(p = -1\) into \(x = h - p\) yields \(x = \frac{1}{2} + 1 = \frac{3}{2}\).

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Key Concepts

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

Rotated Parabola
A rotated parabola is not oriented along the traditional axis-parallel configuration; instead, it is skewed in its direction. This happens when the terms in the equation are not aligned vertically or horizontally. In the equation \(y^2 + 4y + 4x + 2 = 0\), the presence of both the \(y^2\) and linear x-term suggests that the parabola is not aligned with the standard axes. To reroute it back to a more familiar orientation, which makes it easier to analyze and solve, we transform this into a vertex form. Understanding rotated parabolas entails recognizing the need to reformat such equations so the parabola's properties are revealed more clearly.
Complete the Square
Completing the square is a method used to convert a quadratic equation into a form that reveals important features about the graph of the equation. For the terms \(y^2 + 4y\), we complete the square by rearranging them into \((y + 2)^2 - 4\). This step enables us to express the quadratic component in terms of squared expressions. Here's how you complete the square:
  • Take the coefficient of the linear term \(4\), divide it by two to get \(2\), and then square it to obtain \(4\).
  • Add and subtract this square inside the equation to maintain the balance, resulting in the added term \(4\) completing the square.
By completing the square, we simplify the original equation to a form where we can clearly identify the transformations and shifts applied to the original parabola.
Vertex Form of a Parabola
The vertex form of a parabola is incredibly useful because it directly reveals the vertex's position, the parabola's orientation, and the width. For the exercise, we transformed the given equation into \((y+2)^2 = -4(x - \frac{1}{2})\), which is now in vertex form. This transformation tells us several key things:
  • The vertex of the parabola is located at \((\frac{1}{2}, -2)\), translated from the origin due to the terms \(x-\frac{1}{2}\) and \(y+2\).
  • The equation \((y-k)^2 = 4p(x-h)\) indicates that it opens sideways, in this case, to the left, since \(4p\) is negative.
  • The value of \(p\), extracted from the equation \-4p = -4\, is \(-1\), indicating how far the focus is from the vertex along the axis of rotation.
Understanding this form enables the straightforward derivation of other key elements of the parabola, such as the focus and directrix.
Directrix Equation
The directrix is a line associated with every parabola and helps define its geometric properties. For a parabola \((y-k)^2 = 4p(x-h)\), the directrix is given by the equation \(x = h - p\). This formula helps determine the exact location of the directrix relative to the vertex.
  • In this case, substituting \(h = \frac{1}{2}\) and \(p = -1\) into the formula gives us the equation of the directrix as \(x = \frac{1}{2} + 1 = \frac{3}{2}\).
This line represents a constant distance from the vertex, opposite the focus point. The significance of the directrix lies in its role in the definition of a parabola: for any given point on the parabola, the distance to the focus is equal to the distance to the directrix. This focus-directrix property is fundamental in defining the parabolic shape.

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Most popular questions from this chapter

The locus of the vertices of the family of parabolas \(y=\frac{a^{3} x^{2}}{3}+\frac{a^{2} x}{2}-2 a\) is (A) \(x y=\frac{105}{64}\) (B) \(x y=\frac{3}{4}\) (C) \(x y=\frac{35}{16}\) (D) \(x y=\frac{64}{105}\)

Straight lines \(3 x+4 y=5\) and \(4 x-3 y=15\) intersect at A. Points \(B\) and \(C\) are choosen on these lines such that \(A B=A C .\) The equation of the line \(B C\) passing through the point \((1,2)\) is (A) \(x+7 y+13=0\) (B) \(x-7 y+13=0\) (C) \(7 x+y-9=0\) (D) none of these

If the equation of the mirror be \(2 x+y-6=0\) and a ray passing through \((3,10)\) after being reflected by the mirror passes through \((7,2)\), then the equations of the incident ray and the reflected ray are (A) \(x+3 y-13=0\) (B) \(3 x-y+1=0\) (C) \(x-3 y+13=0\) (D) \(3 x+y-1=0\)

Let \(P=(-1,0), Q=(0,0)\) and \(R=(3,3 \sqrt{3})\) be three points. The equation of the bisector of the angle \(P Q R\) (A) \(\sqrt{3} x+y=0\) (B) \(x+\frac{\sqrt{3}}{2} y=0\) (C) \(\frac{\sqrt{3}}{2} x+y=0\) (D) \(x+\sqrt{3} y=0\)

The equation of the straight line passing through the point \((4,3)\) and making intercepts on the co-ordinate axes whose sum is \(-1\) is (A) \(\frac{x}{2}+\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\) (B) \(\frac{x}{2}-\frac{y}{3}=-1\) and \(\frac{x}{-2}+\frac{y}{1}=-1\) (C) \(\frac{x}{2}+\frac{y}{3}=1\) and \(\frac{x}{2}+\frac{y}{1}=1\) (D) \(\frac{x}{2}-\frac{y}{3}=1\) and \(\frac{x}{-2}+\frac{y}{1}=1\)
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