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Chapter 12: Problem 21

Find an equation of the parabola that satisfies the given conditions. vertex \(V(-1,0),\) focus \(F(-4,0)\)

Short Answer

Expert verified
The equation of the parabola is \(y^2 = -12(x + 1)\).

Step by step solution

01

Understand Parabola Orientation

The vertex of the parabola is given as \(V(-1,0)\) and the focus is \(F(-4,0)\). Notice that both these points lie on the line \(y = 0\), confirming the parabola opens horizontally. Since the focus is to the left of the vertex, we know it opens towards the left.
02

Use Vertex-Focus Relation

For a horizontally opening parabola, the equation has the form \[(y - k)^2 = 4p(x - h)\]where \((h,k)\) is the vertex and \(p\) is the directed distance from the vertex to the focus. Here, \(h = -1\), \(k = 0\).
03

Calculate the Distance \(p\)

Calculate \(p\) as the horizontal distance between the vertex \(V(-1,0)\) and the focus \(F(-4,0)\). \[p = -4 - (-1) = -3\]Negative sign indicates the direction towards the focus which is leftwards.
04

Form and Simplify the Parabola Equation

Substitute \(h = -1\), \(k = 0\), and \(p = -3\) into the equation \[(y - 0)^2 = 4(-3)(x + 1)\]Simplify \[y^2 = -12(x + 1)\]This is the equation of the parabola.

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Key Concepts

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

Vertex
The vertex is a key point in understanding parabolas. It is the point where the parabola changes direction. For a parabola in a coordinate plane, it acts as the turning point and provides symmetry to the curve.
In our exercise, the vertex is given as \(V(-1,0)\). This tells us that this particular parabola is horizontally positioned as both the x-coordinate and y-coordinate are aligned horizontally.
  • Vertex \(V(-1,0)\) determines the position and orientation of the parabola.
  • Being on the x-axis, it helps in recognizing the parabola's axis of symmetry.
  • Understanding the vertex's coordinate is crucial for forming the equation of the parabola.
Knowing the vertex location enables one to precisely form the base equation for the parabola's structure.
Focus
The focus is another vital concept when dealing with parabolas. It's a fixed point that, together with the directrix, defines the parabola. In a parabola, every point is equidistant from the focus and the directrix.
In this exercise, the focus is given as \(F(-4, 0)\).
  • The focus being to the left of the vertex indicates the direction of the parabola.
  • It is always located inside the parabola.
  • Understanding where the focus is helps us determine the parabola's opening direction.
This point significantly influences the parabola's equation and leads to understanding the dynamics of its shape.
Horizontally Opening Parabola
A horizontally opening parabola opens on a plane parallel to the x-axis. In contrast to the standard vertical parabola, the vertex and focus align horizontally, dictating its shape and direction.
From the vertex \(V(-1,0)\) and focus \(F(-4,0)\), we conclude this parabola opens horizontally and specifically to the left.
  • Both vertex and focus are along the line \((y=0)\).
  • Direction is determined by the position of focus relative to the vertex.
  • The equation format for a horizontally opening parabola is \( (y - k)^2 = 4p(x - h)\).
Recognizing the parabola's orientation and opening direction helps form the correct equation and understand its geometric properties.
Directed Distance
Directed distance, typically denoted as \(p\), is the measure from the vertex to the focus. In a parabola, it determines how "wide" or "narrow" the parabola appears.
For our exercise, the directed distance is written as \(p = -3\).
  • The negative value indicates the parabola鈥檚 leftward opening.
  • Calculated as \(p = -4 - (-1) = -3\), from the vertex to the focus.
  • Essential for accurately forming the parabola's equation.
Directed distance is an important factor in shaping the parabola in terms of its opening and curvature.
Simplifying Equation
Simplifying the equation is the final step in solving any parabola-related problem. It involves putting all concepts into a cohesive mathematical form. By substituting the known values in the general form of the parabola equation, it expresses the parabola algebraically.
For this horizontally opening parabola:
  • Substitute the vertex \(h = -1\), \(k = 0\).
  • Use directed distance \(p = -3\).
  • The equation becomes \(y^2 = -12(x + 1)\).
This simplification shows the relationship between the terms and presents the equation in a readily usable form, confirming the parabola's properties through its mathematical representation.

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Most popular questions from this chapter

Find an equation for the ellipse that has its center at the origin and satisfies the given conditions. foci \(F(\pm 3,0), \quad\) minor axis of length 2

Exer \(1-16\) : Find the vertices and foci of the conic, and sketch its graph. $$ y^{2}-2 x^{2}+6 y+8 x-3=0 $$

Tangent lines to the parabola \(y=2 x^{2}+3 x+1\) pass through the point \(P(2,-1)\). Find the \(x\) -coordinates of the points of tangency.

(a) Show that the circumference \(C\) of the ellipse with the equation \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) is given by $$ C=4 a \int_{0}^{\pi / 2} \sqrt{1-e^{2} \sin ^{2} \theta} d \theta $$ where \(e\) is the eccentricity. (This is an elliptic integral, which cannot be evaluated using the methods of Chapter 9.) (b) The planet Mercury travels in an elliptical orbit with \(e=0.206\) and \(a=0.387 \mathrm{AU}\). Use part (a) and Simpson's rule, with \(n=10,\) to approximate the length of the orbit. |c) Find the maximum and minimum distances between Mercury and the sun.

The shape of the earth's surface can be approximated by revolving the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1,\) with \(a=\) \(6378 \mathrm{~km}\) and \(b=6356 \mathrm{~km},\) about the \(x\) -axis. Approximate the surface area of the earth to the nearest \(10^{6} \mathrm{~km}^{2}\) (Hint: Use (6.19) with \(f(x)=\sqrt{b^{2}-\left(b^{2} / a^{2}\right) x^{2}},\) and make the substitution \(u=(b / a) x\).)
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