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Chapter 10: Problem 2

Name the conic that has the given equation. Find its vertices and foci, and sketch its graph. $$ y^{2}-6 x=0 $$

Short Answer

Expert verified
The conic is a right-opening parabola with vertex at (0,0) and focus at \((\frac{3}{2}, 0)\).

Step by step solution

01

Identify the Conic Section

The given equation is \(y^2 - 6x = 0\). This is a quadratic equation in one variable (\(y\)) and linear in the other (\(x\)), which suggests that it is a parabola.
02

Rewrite the Equation in Parabola Form

Rearrange the equation to solve for \(x\):\[y^2 = 6x\]This is in the standard form of a parabola that opens horizontally, \(y^2 = 4px\), where \(p\) is the distance from the vertex to the focus.
03

Determine the Vertex and Orientation

For the equation \(y^2 = 6x\), compare it to \(y^2 = 4px\). Here, \(4p = 6\), so \(p = \frac{3}{2}\). This indicates that the parabola opens to the right. The vertex of the parabola is at the origin, \((0,0)\).
04

Find the Focus and Directrix

The focus is \(\frac{3}{2}\) units to the right of the vertex along the x-axis. So, the focus is \((\frac{3}{2}, 0)\). The directrix is \(\frac{3}{2}\) units to the left of the vertex, which is the vertical line \(x = -\frac{3}{2}\).
05

Sketch the Graph

Draw the x and y axes. The vertex is at \((0,0)\), and the parabola opens to the right. Mark the focus at \((\frac{3}{2}, 0)\) and draw a vertical line at \(x = -\frac{3}{2}\) for the directrix. Sketch the parabola opening to the right, centered at the vertex.

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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 stunning geometric curve that is shaped by a quadratic relationship between two variables. In terms of algebra, a parabola's equation typically contains one variable squared, while the other variable remains linear. This is exactly what we find in the equation:
  • \( y^2 - 6x = 0 \)
One unique feature of parabolas is that they can open in different directions. This particular equation, \( y^2 = 6x \), describes a parabola that opens to the right. The reason is the structure \( y^2 = 4px \), where if \( p \) is positive, the parabola opens towards the positive x-direction. It births a wonderful symmetrical curve centered around its vertex.
Vertices and Foci
The vertex of a parabola acts like its center point. It's where the curve changes direction. From this point, you can measure crucial distances that help in sketching the parabola.

For the equation \( y^2 = 6x \):
  • The vertex is at the origin
    \((0,0)\)
  • The focus, which is a point inside the parabola, giving it direction, is calculated by finding \( p \). Here, \( 4p = 6 \), which makes \( p = \frac{3}{2} \)
  • The focus is \( \frac{3}{2} \) units to the right of the vertex, located at \( \left(\frac{3}{2}, 0\right) \)
The directrix of the parabola, often a line outside the curve, sits "opposite" of the focus.
  • For a right-opening parabola like this, the directrix is a vertical line \( x = -\frac{3}{2} \)
Understanding the vertex and focus allows us to grasp not just the parabola鈥檚 location but also its spirit and structure.
Graphing Conics
To bring conic sections like parabolas to life on a graph, start at the basics: the Cartesian plane.
  • First, mark the vertex \((0,0)\) on the graph
  • Then place the focus \( (\frac{3}{2}, 0) \) properly aligned rightward from the vertex
Drawing the parabola:
  • Visualize a curve that bows outwards to the right symmetrically around the x-axis
  • The vertical line \( x = -\frac{3}{2} \) is the directrix
This setup provides a sketch of the parabola, which offers a truly breathtaking view of how algebra transforms into geometry. Each element, from vertex to focus to directrix, melds to shape the smooth curving path of the parabola. This step-by-step plotting on the graph reveals not just the shape but also the grandeur of conic sections.

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Most popular questions from this chapter

Sketch the graph of the given equation, indicating vertices, foci, and asymptotes (if it is a hyperbola). \(4 x^{2}+25 y^{2}=100\)

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Find the points of intersection of \(x^{2}+4 y^{2}=20\) and \(x+2 y=6\)

Find the coordinates of the focus and the equation of the directrix for each parabola. Make a sketch showing the parabola, its focus, and its directrix. \(x^{2}=-16 y\)

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