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Chapter 10: Problem 15

Find the equation of the parabola with vertex at the origin and axis along the \(x\)-axis if the parabola passes through the point \((3,-1)\). Make a sketch.

Short Answer

Expert verified
The equation is \( y^2 = \frac{1}{3}x \).

Step by step solution

01

Understand the Parabola's Equation

For a parabola with its vertex at the origin and its axis along the x-axis, the general equation is of the form \( y^2 = 4px \). Here, \( p \) is the distance from the vertex to the focus. Since the parabola's axis is along the x-axis, \( y \) values depend on \( x \).
02

Use the Given Point to Find p

The parabola passes through the point \((3, -1)\). Substitute these values into the equation \( y^2 = 4px \). This gives:\( (-1)^2 = 4p(3) \).Simplifying, we get:\( 1 = 12p \).
03

Solve for p

Solve the equation from Step 2 for \( p \):\( p = \frac{1}{12} \).
04

Write the Parabola's Equation

Now that we have found \( p \), substitute it back into the equation \( y^2 = 4px \) to get the final equation:\( y^2 = \frac{1}{3}x \).
05

Sketch the Parabola

To sketch the parabola, note that it opens to the right along the x-axis because the equation is in the form where \( y^2 \) determines \( x \). The vertex is at the origin \((0, 0)\), and it passes through the point \((3, -1)\). The parabola is relatively flat due to the small value of \( p \).

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Key Concepts

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

Vertex of a Parabola
The vertex of a parabola is a critical point that often serves as a starting point in understanding its geometry and equation. When dealing with parabolas, the vertex is the point where the curve changes direction. For some, it's seen as the 'tip' of the parabola.In the context of the given problem, the vertex is conveniently located at the origin, (0, 0). This positioning makes it easier to work with the standard form equation of a parabola,
  • For parabolas aligned along the x-axis: \(y^2 = 4px\)
  • For parabolas aligned along the y-axis: \(x^2 = 4py\)
In our exercise, having the vertex at the origin and the given point (3, -1) was essential in finding the precise equation of the parabola. Remember, the vertex isn't just a point; it's the heart of the parabola where most calculations begin.
Axis of Symmetry
The axis of symmetry in a parabola is a straight line that divides the curve into two identical halves. Imagine it as a mirror line where one side of the parabola reflects onto the other. Understanding this line helps in sketching the parabola accurately and comprehensively.For the parabola in the exercise, the axis of symmetry is along the x-axis. This means the parabola opens horizontally. This characteristic of symmetry means that reflecting the points across the x-axis yields the rest of the parabola.A clear hint that a parabola has its axis along the x-axis in an equation is when the equation is in the form \(y^2 = 4px\). The symmetry of a parabola makes setting it on a graph more predictable. Knowing the axis of symmetry can guide you not only mathematically but visually.
Focus of a Parabola
The focus of a parabola is a special point located inside the curve. It's the point around which the parabola is 'dragged'. Any point on the parabola is equidistant to the focus and a line called the directrix.In this exercise, once you've determined that the equation is \(y^2 = 4px\) and you found \(p = \frac{1}{12}\), you can conclude that the focus lies at the point (\(\frac{1}{12}\), 0). This implies that the focus is very close to the origin, guiding us to the next key property.
  • The formula \(4p\) encompasses the geometric properties of the parabola.
  • If \(p\) is positive, the parabola opens rightwards; if negative, leftwards.
  • The closer the focus, the steeper the parabola's opening.
Knowing the focus assists in understanding the depth and direction of the parabola鈥檚 curve. It keeps the relationship between all elements of the parabola coherent and predictable.

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Most popular questions from this chapter

Plot the points whose polar coordinates follow. For each point, give four other pairs of polar coordinates, two with positive \(r\) and two with negative \(r\). (a) \(\left(1, \frac{1}{2} \pi\right)\) (b) \(\left(-1, \frac{1}{4} \pi\right)\) (c) \(\left(\sqrt{2},-\frac{1}{3} \pi\right)\) (d) \(\left(-\sqrt{2}, \frac{5}{2} \pi\right)\)

The position of a comet with a highly eccentric elliptical orbit ( \(e\) very near 1 ) is measured with respect to a fixed polar axis (sun is at a focus but the polar axis is not an axis of the ellipse) at two times, giving the two points \((4, \pi / 2)\) and \((3, \pi / 4)\) of the orbit. Here distances are measured in astronomical units ( \(1 \mathrm{AU} \approx 93\) million miles). For the part of the orbit near the sun, assume that \(e=1\), so the orbit is given by $$ r=\frac{d}{1+\cos \left(\theta-\theta_{0}\right)} $$ (a) The two points give two conditions for \(d\) and \(\theta_{0}\). Use them to show that \(4.24 \cos \theta_{0}-3.76 \sin \theta_{0}-2=0\). (b) Solve for \(\theta_{0}\) using Newton's Method. (c) How close does the comet get to the sun?

Plot the following parametric curves. Describe in words how the point moves around the curve in each case. (a) \(x=\cos \left(t^{2}-t\right), y=\sin \left(t^{2}-t\right)\) (b) \(x=\cos \left(2 t^{2}+3 t+1\right), y=\sin \left(2 t^{2}+3 t+1\right)\) (c) \(x=\cos (-2 \ln t), y=\sin (-2 \ln t)\) (d) \(x=\cos (\sin t), y=\sin (\sin t)\)

Using the same axes, draw the conics \(y=\) \(\pm\left(a x^{2}+1\right)^{1 / 2}\) for \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\) using \(a=\) \(-2,-1,-0.5,-0.1,0,0.1,0.6,1\). Make a conjecture about how the shape of the figure depends on \(a\).

Find the area of the ellipse \(b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}\).
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