Parabolas

Section 6.7 Parabolas

A parabola is a curve that is defined by a special geometric property. Every point on a parabola is the same distance from a special point called its focus and a special line called its directrix. In the figure below the focus is labeled \(F\) and the directrix is labeled \(l\text{.}\) The parabola is the “U”-shaped curve. Observe the equal distances to the focus and directrix at the two points illustrated on the curve. The point \(V\) at the base of “U”-shape is called the vertex of the parabola.

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Example 6.7.1.

Let’s find an equation for the parabola with focus at \(F(0,1)\) and directrix \(y = -1\text{.}\)

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Suppose \(P(x,y)\) is a point on the parabola. Then the distance from \(P\) to \(F\) is given by the Distance Formula ((6.1.1)).

\begin{equation*} \overline{PF} = \sqrt{(x-0)^2+(y-1)^2} = \sqrt{x^2+(y-1)^2}. \end{equation*}

The distance from \(P\) to the line \(y = -1\) is \(\overline{Pl} = y+1\text{.}\) These distances must be equal so that

\begin{equation*} \sqrt{x^2+(y-1)^2} = y+1. \end{equation*}

Squaring both sides gives

\begin{equation*} x^2+(y-1)^2 = (y+1)^2. \end{equation*}

Expanding the squared terms

\begin{equation*} x^2+y^2-2y+1 = y^2+2y+1 \end{equation*}

Canceling common terms from both sides:

\begin{equation*} x^2-2y = 2y \end{equation*}

And combining like terms:

\begin{equation*} x^2 = 4y \end{equation*}

So that

\begin{equation*} y = \frac{1}{4} x^2 \end{equation*}

is an equation for our parabola.

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Activity 6.7.1. Parabolas and Quadratic Equations.

In fact, a curve whose equation is given by a quadratic polynomial of the form

\begin{equation*} y = ax^2+bx+c\text{,} \end{equation*}

where \(a\neq 0\) has the graph of a parabola. In the demonstration below, move the sliders to consider the effect of each coefficient.

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Figure 6.7.2. Graph of a parabola \(y=ax^2+bx+c\text{.}\)

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(a)

Adjust the slider for the constant coefficient \(c\text{.}\) What effect does it have on the graph?

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(b)

Adjust the slider for the quadratic coefficient \(a\text{.}\) What effect does it have on the graph? Make sure to experiment with positive and negative values. What happens if \(a=0\text{?}\)

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(c)

Adjust the slider for the linear coefficient \(b\text{.}\) What effect does it have on the graph?

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Example 6.7.3.

Graph the parabola \(y = -2 x^2+5 x+3\) by finding the intercepts of the graph.

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Solution.

To find where the graph intersects the \(x\)-axis, we need to set \(y=0\) and solve for \(x\text{.}\) This can be done, in this case, via factoring:

\begin{align*} -2 x^2+5 x+3 \amp= 0\\ -(2 x^2-5 x-3 \amp= 0\\ -(2x+1)(x-3) \amp= 0 \end{align*}

So that \(2x+1=0\) OR \(x-3 = 0\text{.}\) This gives \(x\)-intercepts at \((-1/2,0)\) and \((3,0)\text{.}\)

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To find where the graph intersects the \(y\)-axis, we need to evaluate at \(x=0\text{.}\) We have

\begin{equation*} y = -2 (0)^2+5(0)+3 = 3, \end{equation*}

so that \((0,3)\) is the \(y\)-intercept. Since the leading coefficient is negative, the parabola will open down.

The vertex can be found equidistant from the two \(x\)-intercepts. To find its \(x\)-coordinate, average

\begin{equation*} \frac{-\frac{1}{2}+3}{2} = \frac{-1+6}{4} = \frac{5}{4}. \end{equation*}

To find the \(y\)-coordinate of the vertex, evaluate

\begin{align*} y \amp= -2 \left(\frac{5}{4}\right)^2 + 5 \left(\frac{5}{4}\right)+ 3\\ \amp= -2 \left(\frac{25}{16}\right) + \frac{25}{4}+ 3\\ \amp= -\frac{50}{16} + \frac{25}{4}+ 3\\ \amp= -\frac{25}{8} + \frac{50}{8}+ \frac{24}{8}\\ \amp= \frac{-25+50+24}{8} \\ \amp= \frac{49}{8} \end{align*}

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