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Chapter 10: Problem 15

Find an equation for the parabola that satisfies the given conditions. $$ \begin{array}{l}{\text { (a) Vertex }(0,0) ; \text { focus }(3,0)} \\ {\text { (b) Vertex }(0,0) ; \text { directrix } y=\frac{1}{4}}\end{array} $$

Short Answer

Expert verified
(a) The equation is \( y^2 = 12x \). (b) The equation is \( x^2 = -y \).

Step by step solution

01

Understanding Parabola Parts

A parabola is defined by its vertex, focus, and directrix. For option (a), we have a vertex at \(0,0\) and a focus at \(3,0\). For option (b), the vertex is \(0,0\) and the directrix is \(y = \frac{1}{4}\). Both cases involve parabolas with horizontal axes due to the given points and lines.
02

Choosing Parabola Formula for (a)

For a parabola with a horizontal axis centered at the origin, the standard form of the equation is \(y^2 = 4px\). The vertex is \(0,0\) and the focus has coordinates \(p,0\). Given the focus \(3,0\), we determine \(p = 3\). Therefore, the equation becomes \(y^2 = 12x\).
03

Finding Equation for (a)

Substitute \(p = 3\) into \(y^2 = 4px\) to make the equation \(y^2 = 12x\). This equation describes a parabola with a vertex at the origin and a focus at \(3,0\).
04

Choosing Parabola Formula for (b)

For part (b) with a vertical axis and vertex at the origin, the standard form is \(x^2 = 4py\), where the directrix is given by \(y = -p\). Since the directrix provided in part (b) is \(y = \frac{1}{4}\), we equate \( -p = \frac{1}{4}\), so \(p = -\frac{1}{4}\). This means the equation is \(x^2 = -y\).
05

Finding Equation for (b)

Substitute \(p = -\frac{1}{4}\) into \(x^2 = 4py\) making \(x^2 = -y\). Therefore, the parabola opens downwards with the vertex at the origin and the specified directrix.

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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 where the parabola changes direction. It represents the tip or the lowest/highest point of the curve, depending on its orientation. In our exercise, the vertex is at (0,0) for both scenarios.
  • For horizontal parabolas, the vertex visually sits at the middle of the curve horizontally.
  • For vertical parabolas, the vertex is the middle of the curve vertically.
To find the vertex's position in equations, one can identify it from the standard form expressions, either in the form \( y^2 = 4px \) for horizontal parabolas or \( x^2 = 4py \) for vertical parabolas. The position of the vertex helps determine the entire layout of the parabola.
Focus
The focus of a parabola is a point located inside its curve. It plays a crucial role in defining the parabola's shape. In our examples, one parabola has a focus at (3,0). Here鈥檚 a closer look:
  • The focus serves as a point where all incoming lines of symmetry meet or seem to be directed towards it.
  • It is always located inside the parabola, directly along the axis that the parabola opens towards.
  • In the equation \( y^2 = 4px \), 'p' defines the distance between the vertex and the focus. For instance, a focus at (3,0) means \( p = 3 \).
Understanding where the focus is located gives insight into how wide the parabola opens.
Directrix
The directrix is a straight line that helps shape a parabola alongside its focus. For part (b) of the exercise, the directrix is given by \( y = \frac{1}{4} \).Here's what you need to know:
  • The directrix is positioned outside of the curve, contrary to the focus which lies within.
  • It is perpendicular to the axis of symmetry of the parabola.
  • For vertical parabolas such as \( x^2 = 4py \), the directrix can be expressed as \( y = -p \).
  • In our case, \( y = \frac{1}{4} \) for part (b), which indicates \( p = -\frac{1}{4} \).
The directrix, together with the focus, defines how the parabola hugs its main axis.
Standard Form of Parabola
The standard form of a parabola's equation dictates its specific curvature and orientation. Depending on whether the axis is horizontal or vertical, the form adjusts:
  • A horizontal axis relies on the form \( y^2 = 4px \). It opens rightward or leftward from the vertex.
  • A vertical axis uses \( x^2 = 4py \). It presents upward or downward openings from the vertex.
For scenario (a), with a focus of (3,0), the standard form becomes \( y^2 = 12x \) by calculating \( 4p = 12 \). In scenario (b), using the directrix at \( y = \frac{1}{4} \), we derived the equation \( x^2 = -y \).
The standard forms are pivotal as they elucidate how wide or narrow a parabola will be, and which direction it opens.

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Most popular questions from this chapter

Find the eccentricity and the distance from the pole to the directrix, and sketch the graph in polar coordinates. $$ \text { (a) } r=\frac{4}{2+3 \cos \theta} \quad \text { (b) } r=\frac{5}{3+3 \sin \theta} $$

Prove: The line tangent to the parabola \(x^{2}=4 p y\) at the point \(\left(x_{0}, y_{0}\right)\) is \(x_{0} x=2 p\left(y+y_{0}\right)\)

Sketch the hyperbola, and label the vertices, foci, and asymptotes. $$ \begin{array}{l}{\text { (a) } x^{2}-4 y^{2}+2 x+8 y-7=0} \\ {\text { (b) } 16 x^{2}-y^{2}-32 x-6 y=57}\end{array} $$

True鈥揊alse Determine whether the statement is true or false. Explain your answer. $$ \begin{array}{l}{\text { The } x \text { -axis is tangent to the polar curve } r=\cos (\theta / 2) \text { at }} \\ {\theta=3 \pi .}\end{array} $$

True鈥揊alse Determine whether the statement is true or false. Explain your answer. If an ellipse is not a circle, then the eccentricity of the ellipse is less than one.
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