7.2: Parabolas

Like ellipses, you have seen parabolas (e.g. \(y=x^2\)) and some of their applications (e.g. projectile trajectories), but perhaps without knowing their purely geometric definition. The alternative definition of an ellipse described in Exercise [exer:ellipdirectrix] in Section 7.1 is, in fact, similar to the definition of the parabola:

Figure [fig:parabolavert] illustrates the above definition, with a point \(P\) moving along the parabola so that the distance from \(P\) to the focus \(F\) equals the distance from \(P\) to the directrix \(D\). Note that the point halfway between the focus and directrix must be on the parabola—that point is the vertex, which is the point on the parabola closest to the directrix. The axis of the parabola is the line that passes through the focus and is perpendicular to the directrix. Notice that the ratio \(\frac{PF}{PG}\) equals 1, whereas that ratio for an ellipse—by the alternative definition—was the eccentricity \(e<1\). The eccentricity of the parabola, therefore, is always 1.⁴

To construct a parabola from the definition, cut a piece of string to have the same length \(AB\) as one side of a drafting triangle, as in Figure [fig:paraboladraw].

Fasten one end of the string to the vertex \(A\) of the triangle and the other end to a pin somewhere between \(A\) and \(B\)—the pin will be the focus \(F\) of the parabola. Hold the string taut against the edge \(\overline{AB}\) of the triangle at a point \(P\) on either side of the pin, then move the edge \(\overline{BC}\) of the triangle along the directrix \(D\). The drawn figure will be a parabola, as the lengths \(PF\) and \(PB\) will be equal (since the length of the string being \(AB=AP+PF\) means \(PF=PB\)). To derive the equation of a parabola in the \(xy\)-plane, start with the simple case of the focus on the \(y\)-axis at \((0,p)\), with \(p>0\), and the line \(y=-p\) as the directrix, as in the figure on the right. The vertex is then at the origin \((0,0)\). Pick a point \((x,y)\) whose distances \(d_1\) and \(d_2\) from the focus \((0,p)\) and directrix \(y=-p\), respectively, are equal. Then

\[\begin{aligned} d_1^2 ~&=~ d_2^2\\ (x-0)^2 ~+~ (y-p)^2 ~&=~ (x-x)^2 ~+~ (y+p)^2\\ x^2 ~+~ \cancel{y^2} ~-~ 2py ~+~ \cancel{p^2} ~&=~ \cancel{y^2} ~+~ 2py ~+~ \cancel{p^2}\\ x^2 ~&=~ 4py\end{aligned} \nonumber \]

In other words, \(y = \frac{1}{4p}x^2\), which is the more familiar form of a parabola. Thus, any curve of the form \(y=ax^2\), with \(a \ne 0\), is a parabola whose focus and directrix can be found by dividing \(a\) by \(4\): \(p = \frac{a}{4}\), so that the focus is at \(\left(0,\frac{a}{4}\right)\) and the directrix is the line \(y=-\frac{a}{4}\). For example, the parabola \(y=x^2\) has its focus at \(\left(0,\frac{1}{4}\right)\) and its directrix is the line \(y=-\frac{1}{4}\).

When \(p>0\) the parabola \(4py = x^2\) extends upward; for \(p<0\) it extends downward, as in Figure [fig:parabolap](a) below:

Switching the roles of \(x\) and \(y\) yields the parabola \(4px=y^2\), with focus at \((p,0)\) and directrix \(x=-p\). For \(p>0\) this parabola extends rightward, while for \(p<0\) it extends leftward. See Figure [fig:parabolap](b) and (c).

It is left as an exercise to show that in general a curve of the form \(y=ax^2+bx+c\) is a parabola. Just like not every oval shape is an ellipse, not every “cupped” or “U” shape is a parabola (e.g. \(y=x^4\)). The slope of the parabola \(4py=x^2\) is \(\dydx=\frac{2x}{4p}=\frac{x}{2p}\), so that the equation of the tangent line to the parabola at a point \((x_0,y_0)\) is:

\[\begin{aligned} y ~-~ y_0 ~&=~ \frac{x_0}{2p}\,(x - x_0)\nonumber\\ 2p\,(y-y_0) ~&=~ x_0x ~-~ x_0^2\nonumber\\ 2py ~-~ 2py_0 ~&=~ x_0x ~-~ 4py_0\nonumber\\ 2p\,(y+y_0) ~&=~ x_0x\label{eqn:parabtangenty}\end{aligned} \]

Likewise, switching the roles of \(x\) and \(y\), the tangent line to the parabola \(4px=y^2\) at a point \((x_0,y_0)\) is:

\[\label{eqn:parabtangentx} 2p\,(x+x_0) ~=~ y_0y \]

Formula ([eqn:parabtangentx]) simplifies the proof of the reflection property for parabolas: light shone from the focus to any point on the parabola will reflect in a path parallel to the axis of the parabola. Figure [fig:parabreflect] shows light emanating from the focus \(F=(p,0)\) and reflecting off a point \(P=(x_0,y_0)\) on the parabola \(4px=y^2\). If that line of reflection is parallel to the \(x\)-axis—the axis of the parabola—then the tangent line to the parabola at \((x_0,y_0)\) should make the same angle \(\beta\) with the line of reflection as it does with the \(x\)-axis. So extend the tangent line to intersect the \(x\)-axis and use formula ([eqn:parabtangentx]) to find the \(x\)-intercept:

\[2p\,(x+x_0) ~=~ y_0y ~=~ y_0 \cdot 0 ~=~ 0 \quad\Rightarrow\quad x ~=~ -x_0 \nonumber \]

Let \(Q=(-x_0,0)\), so that the distance \(FQ\) equals \(p+x_0\). The goal is to show that the angle of incidence \(\angle FPQ\) equals the angle of reflection \(\beta\). The focal radius \(\overline{FP}\) has length

\[FP ~=~ \sqrt{(p-x_0)^2 + (0-y_0)^2} ~=~ \sqrt{p^2 - 2px_0 + x_0^2 + 4px_0} ~=~ \sqrt{p^2 + 2px_0 + x_0^2} ~=~ p+x_0 ~. \nonumber \]

Thus, \(FQ=FP\) in the triangle \(\triangle FPQ\), so that \(\angle FPQ = \angle FQP = \beta\), i.e. the light’s path does indeed satisfy Fermat’s Principle for curved surfaces.\(\quad\checkmark\)

The parabola’s reflection property shows up in some engineering applications, typically by revolving part of a parabola around its axis, producing a parabolic surface in three dimensions called a paraboloid. For example, it used to be common for vehicle headlights to use paraboloids for their inner reflective surface, with a bulb at the focus, so that—by the reflection property—the light would shine straight ahead in a solid beam. Many flashlights still operate on that principle. The reflection property also works in the opposite direction, which is why satellite dishes and radio telescopes are often wide paraboloids with a signal receiver at the focus, to maximize reception of incoming reflected signals.

Example \(\PageIndex{1}\): parabenvelope

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Solution

Suppose that an object is launched from the ground with an initial velocity \(v_0\) and at varying angles with the ground. Show that the family of all the possible trajectories—which are parabolic—form a region whose boundary (called the envelope of the trajectories) is itself a parabola.

Solution: Recall from Example

Example \(\PageIndex{1}\): minmax3

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Solution

in Section 4.1 that if the object is launched at an angle \(0<\theta<\frac{\pi}{2}\) with the ground, then the height \(y\) attained by the object as a function of the horizontal distance \(x\) that it travels is given by

\[y ~=~ -\frac{gx^2}{2v_0^2 \cos^2\,\theta} ~+~ x\tan\,\theta ~. \nonumber \]

The curve is a parabola, with the figure on the right showing these parabolic trajectories for 500 values of the angle \(\theta\). Clearly each parabola intersects at least one other.The maximum horizontal distance \(\frac{v_0^2}{g}\) occurs only for \(\theta=\frac{\pi}{4}\), as was shown in Example

Example \(\PageIndex{1}\): minmax3

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Solution

. The maximum vertical height \(\frac{v_0^2}{2g}\) is attained when the object is launched straight up (i.e. \(\theta = \frac{\pi}{2}\)), as was shown in Exercise [exer:projmax0] in Section 5.1. By symmetry only the angles \(0<\theta\le\frac{\pi}{2}\) in the same vertical plane need be considered. So in the above figure, imagine if the trajectories for all possible angles were included, filling up a region that does appear to have a parabolic boundary. This will now be shown to be true.

First, it turns out that all the parabolas for \(0<\theta<\frac{\pi}{2}\) have the same directrix \(y=\frac{v_0^2}{2g}\). To see why, recall from Exercise [exer:projmaxangle] in Section 4.1 that the maximum height reached by the object is \(\frac{v_0^2\,\sin^2 \theta}{2g}\), which is thus the \(y\)-coordinate of the vertex of the parabola. That vertex is midway between the focus and the directrix. The parabola is of the form \(4py = x^2 + bx\), where \(b\) is a constant that does not affect the distance between the vertex and directrix⁵, and \(4p\) is a constant with \(p<0\) such that the directrix is \(-p\) units above the vertex (since \(p<0\)), just as in the case \(4py=x^2\). The equation of the parabola then shows that

\[\frac{1}{4p} ~=~ -\frac{g}{2v_0^2 \cos^2\,\theta} \quad\Rightarrow\quad p ~=~ -\frac{v_0^2 \cos^2\,\theta}{2g} \nonumber \]

so that the directrix is at

\[\begin{aligned} y ~&=~ \text{$y$-coordinate of the vertex} ~+~ (-p)\\ &=~ \frac{v_0^2\,\sin^2 \theta}{2g} ~+~ -\left(-\frac{v_0^2 \cos^2\,\theta}{2g}\right) ~=~ \frac{v_0^2}{2g}(\sin^2 \theta ~+~ \cos^2\,\theta)\\ y ~&=~ \frac{v_0^2}{2g}\end{aligned} \nonumber \]

It is perhaps surprising that all the parabolic trajectories share the same directrix \(y=\frac{v_0^2}{2g}\), which is independent of the angle \(\theta\). Note that the heights of each vertex \(\left(\frac{v_0^2\,\sin^2 \theta}{2g}\right)\) and focus \(\left(\frac{v_0^2}{2g}(\sin^2 \theta - \cos^2\,\theta)\right)\) do depend on \(\theta\). The common directrix is the key to the remainder of the proof. Now let \(P\) be a point in the first quadrant of the \(xy\)-plane below the common directrix \(y=\frac{v_0^2}{2g}\), denoted by \(D\). Then \(P\) can be either inside, outside, or on the envelope, as in Figure [fig:envelope3]:

The origin \(O=(0,0)\) is on each trajectory, so by definition of a parabola the foci for all the trajectories must be a distance \(\frac{v_0^2}{2g}\) from \(O\), i.e. the distance from \(O\) to \(D\). That is, the foci of all the trajectories must lie on the circle \(C_0\) of radius \(\frac{v_0^2}{2g}\) centered at \(O\). If \(P\) is any other point inside the envelope, so that it lies on at least one trajectory, then it must be a distance \(r>0\) below the line \(D\). By definition of a parabola, \(P\) must be the same distance from the foci of any trajectories it belongs to. That is, the foci must be on a circle \(C\) of radius \(r\) centered at \(P\) and touching the directrix \(D\), as in Figure [fig:envfoci3]:

In Figure [fig:envfoci3](a) \(C\) and \(C_0\) intersect at two points \(F_1\) and \(F_2\), so \(P\) belongs to two trajectories; \(P\) must then be inside the envelope. In Figure [fig:envfoci3](b) \(C\) and \(C_0\) do not intersect, so \(P\) must be outside the envelope (since it is not on a parabola with a focus on \(C_0\)). If \(C\) and \(C_0\) intersect at only one point \(F\), as in Figure [fig:envfoci3](c), then \(P\) must be on the envelope. In that case, \(P\) is a distance \(r+\frac{v_0^2}{2g}\) from \(O\), which is also the distance from \(P\) to the line \(y=\frac{v_0^2}{g}\) (denoted by \(L\)). Thus, by definition of a parabola, \(P\) is on a parabola with focus \(O\) and directrix \(L\). The vertex is at \(\left(0,\frac{v_0^2}{2g}\right)\). Therefore, the envelope is a parabola: the boundary of the shaded region in the figure on the right.\(\quad\checkmark\)

In Example

Example \(\PageIndex{1}\): parabenvelope

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Solution

all the trajectories were in the \(xy\)-plane only. Removing that restriction, so that trajectories in all vertical planes through the \(y\)-axis are possible, would result in a solid paraboloid consisting of all possible trajectories from the origin. Parabolas also appear in suspension bridges: the suspension cables supporting a horizontal bridge (via vertical suspenders, as in the figure on the right) have to be parabolas if the weight of the bridge is uniformly distributed.⁶

[sec7dot2]

Construct a parabola using the procedure shown in Figure [fig:paraboladraw].

For Exercises 2-6, sketch the graph of the given parabola and indicate the exact locations of the focus, vertex, and directrix. [[1.]]

5

\(8y=x^2\)

\(y=8x^2\)

\(x=y^2\)

\(x=-3y^2\)

\(-1000y=x^2\)

Find the points of intersection of the parabolas \(4py=x^2\) and \(4px=y^2\) when \(p>0\). What is the equation of the line through those points?

A vehicle headlight in the shape of a paraboloid is 3in deep and has an open edge with diameter 8in. Where should the center of the bulb be placed in order to be at the focus, measured in inches relative to the vertex?

The latus rectum of a parabola is the chord that passes through the focus and is parallel to the directrix. Find the length of the latus rectum for the parabola \(4py=x^2\).

Show that the circle whose diameter is the latus rectum of a parabola touches the parabola’s directrix at one point.

Find the points on the parabola \(4px=y^2\) such that the focal radii to those points have the same length as the latus rectum.

From each end of the latus rectum of a parabola draw a line to the point where the directrix and axis intersect. Show that the two drawn lines are perpendicular. [[1.]]

Show that any point not on a parabola is on either zero or two tangent lines to the parabola.

Show that \(y=mx-2mp-m^3p\) is the normal line of slope \(m\) to the parabola \(4px=y^2\).

From a point \(P\) on a parabola with vertex \(V\) let \(\overline{PQ}\) be the line segment perpendicular to the axis at a point \(Q\). Show that \(PQ^2\) equals the product of \(QV\) and the length of the latus rectum.

Show that the curve \(y=ax^2+bx+c\) is a parabola for \(a \ne 0\), using only the definition of a parabola. Find the focus, vertex and directrix.

Show that the set of all midpoints of a family of parallel chords in a parabola lie on a line parallel to the parabola’s axis.