Eccentricity Of A Conic Section

Learning Objectives

• 7.5.i Place the equation of a parabola in standard form with given focus and directrix.
• seven.five.ii Identify the equation of an ellipse in standard form with given foci.
• 7.5.iii Identify the equation of a hyperbola in standard course with given foci.
• 7.5.4 Recognize a parabola, ellipse, or hyperbola from its eccentricity value.
• 7.5.5 Write the polar equation of a conic section with eccentricity $e$.
• 7.5.six Identify when a full general equation of degree ii is a parabola, ellipse, or hyperbola.

Conic sections accept been studied since the time of the aboriginal Greeks, and were considered to be an of import mathematical concept. Equally early on equally 320 BCE, such Greek mathematicians as Menaechmus, Appollonius, and Archimedes were fascinated by these curves. Appollonius wrote an entire 8-volume treatise on conic sections in which he was, for example, able to derive a specific method for identifying a conic section through the use of geometry. Since so, important applications of conic sections have arisen (for example, in astronomy), and the properties of conic sections are used in radio telescopes, satellite dish receivers, and even compages. In this section we discuss the three basic conic sections, some of their properties, and their equations.

Conic sections go their name because they can be generated by intersecting a plane with a cone. A cone has two identically shaped parts called nappes. One nappe is what near people hateful by "cone," having the shape of a party hat. A correct circular cone can exist generated past revolving a line passing through the origin around the y-axis as shown.

Figure 7.43 A cone generated by revolving the line $y=3\mathrm{ten}$ around the $y$-axis.

Conic sections are generated by the intersection of a plane with a cone (Figure 7.44). If the plane intersects both nappes, then the conic section is a hyperbola. If the aeroplane is parallel to the generating line, the conic section is a parabola. If the plane is perpendicular to the axis of revolution, the conic section is a circle. If the plane intersects one nappe at an angle to the axis (other than $\mathrm{xc}\text{°}),$ so the conic department is an ellipse.

Effigy 7.44 The 4 conic sections. Each conic is determined by the angle the airplane makes with the axis of the cone.

Parabolas

A parabola is generated when a aeroplane intersects a cone parallel to the generating line. In this case, the plane intersects simply one of the nappes. A parabola can also exist defined in terms of distances.

Definition

A parabola is the set up of all points whose distance from a fixed point, called the focus, is equal to the altitude from a fixed line, called the directrix. The point halfway between the focus and the directrix is called the vertex of the parabola.

A graph of a typical parabola appears in Effigy seven.45. Using this diagram in conjunction with the distance formula, we can derive an equation for a parabola. Recall the distance formula: Given point P with coordinates $\left({\mathrm{10}}_1,y_{\mathrm{one}}\right)$ and point Q with coordinates $\left({\mathrm{ten}}_2,{\text{y}}_2\right),$ the distance between them is given past the formula

$$d\left(P,Q\right)=\sqrt{{\left(x_{\mathrm{ii}}-x_{\mathrm{i}}\right)}^2+{\left(y_2-y_{\mathrm{ane}}\right)}^2}.$$

Then from the definition of a parabola and Figure 7.45, we get

$$\begin{array}{ccc}\hfill d\left(F,P\right)& =\hfill & d\left(P,Q\right)\hfill \\ \hfill \sqrt{{\left(0-x\right)}^2+{\left(p-y\right)}^{\mathrm{ii}}}& =\hfill & \sqrt{{\left(x-\mathrm{ten}\right)}^{\mathrm{ii}}+{\left(\text{−}p-y\right)}^2}.\hfill \end{array}$$

Squaring both sides and simplifying yields

$$\begin{array}{ccc}\hfill x^2+{\left(p-y\right)}^2& =\hfill & 0^2+{\left(\text{−}p-y\right)}^2\hfill \\ \hfill x^{\mathrm{ii}}+p^2-2py+y^{\mathrm{ii}}& =\hfill & p^2+2py+y^{\mathrm{two}}\hfill \\ \hfill {\mathrm{10}}^{\mathrm{two}}-\mathrm{two}py& =\hfill & \mathrm{two}py\hfill \\ \hfill x^2& =\hfill & 4py.\hfill \end{array}$$

Figure 7.45 A typical parabola in which the distance from the focus to the vertex is represented past the variable $p.$

Now suppose we desire to relocate the vertex. We use the variables $\left(h,g\right)$ to announce the coordinates of the vertex. And so if the focus is directly to a higher place the vertex, information technology has coordinates $\left(h,k+p\right)$ and the directrix has the equation $y=k-p.$ Going through the same derivation yields the formula ${\left(x-h\right)}^2=4p\left(y-\mathrm{thousand}\right).$ Solving this equation for y leads to the following theorem.

Theorem 7.viii

Equations for Parabolas

Given a parabola opening upward with vertex located at $\left(h,\mathrm{thousand}\right)$ and focus located at $\left(h,k+p\right),$ where p is a constant, the equation for the parabola is given by

$$y=\frac{1}{4p}{\left(x-h\right)}^2+k.$$

(7.11)

This is the standard class of a parabola.

We can as well study the cases when the parabola opens downwardly or to the left or the right. The equation for each of these cases tin too exist written in standard course equally shown in the following graphs.

Figure 7.46 Four parabolas, opening in diverse directions, along with their equations in standard grade.

In addition, the equation of a parabola tin can be written in the general course, though in this grade the values of h, k, and p are non immediately recognizable. The general grade of a parabola is written every bit

$$a{x}^2+bx+cy+d=0\text{or}a{y}^{\mathrm{two}}+b\mathrm{ten}+cy+d=0.$$

The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard class, utilize the method of completing the square.

Example 7.xix

Converting the Equation of a Parabola from Full general into Standard Form

Put the equation $x^2-4x-8y+12=0$ into standard grade and graph the resulting parabola.

Checkpoint vii.18

Put the equation $2y^2-x+12y+\mathrm{sixteen}=0$ into standard form and graph the resulting parabola.

The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective holding. Suppose we accept a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as low-cal or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

Consider a parabolic dish designed to collect signals from a satellite in infinite. The dish is aimed directly at the satellite, and a receiver is located at the focus of the parabola. Radio waves coming in from the satellite are reflected off the surface of the parabola to the receiver, which collects and decodes the digital signals. This allows a minor receiver to assemble signals from a wide angle of sky. Flashlights and headlights in a car piece of work on the same principle, merely in contrary: the source of the lite (that is, the calorie-free bulb) is located at the focus and the reflecting surface on the parabolic mirror focuses the beam directly ahead. This allows a pocket-size light bulb to illuminate a wide angle of space in front of the flashlight or car.

Ellipses

An ellipse can also be defined in terms of distances. In the case of an ellipse, there are 2 foci (plural of focus), and two directrices (plural of directrix). We look at the directrices in more than detail after in this section.

Definition

An ellipse is the fix of all points for which the sum of their distances from two fixed points (the foci) is abiding.

Effigy 7.48 A typical ellipse in which the sum of the distances from any point on the ellipse to the foci is constant.

A graph of a typical ellipse is shown in Effigy 7.48. In this figure the foci are labeled equally $F$ and $F^{\prime }.$ Both are the aforementioned fixed distance from the origin, and this distance is represented by the variable c. Therefore the coordinates of $F$ are $\left(c,0\right)$ and the coordinates of $F^{\prime }$ are $\left(\text{−}c,0\right).$ The points $P$ and $P^{\prime }$ are located at the ends of the major axis of the ellipse, and have coordinates $\left(a,0\right)$ and $\left(\text{−}a,0\right),$ respectively. The major axis is always the longest distance across the ellipse, and can exist horizontal or vertical. Thus, the length of the major axis in this ellipse is 2a. Furthermore, $P$ and $P^{\prime }$ are chosen the vertices of the ellipse. The points $Q$ and $Q^{\prime }$ are located at the ends of the pocket-sized axis of the ellipse, and have coordinates $\left(0,b\right)$ and $\left(0,\text{−}b\right),$ respectively. The minor centrality is the shortest distance across the ellipse. The minor axis is perpendicular to the major axis.

Co-ordinate to the definition of the ellipse, we can cull whatsoever point on the ellipse and the sum of the distances from this point to the two foci is constant. Suppose we choose the point P. Since the coordinates of point P are $\left(a,0\right),$ the sum of the distances is

$$d\left(P,F\right)+d\left(P,F^{\prime }\right)=\left(a-c\right)+\left(a+c\right)=\mathrm{ii}a.$$

Therefore the sum of the distances from an arbitrary point A with coordinates $\left(x,y\right)$ is besides equal to twoa. Using the distance formula, we get

$$\begin{array}{ccc}\hfill d\left(A,F\right)+d\left(A,F^{\prime }\right)& =\hfill & 2a\hfill \\ \hfill \sqrt{{\left(x-c\right)}^2+y^{\mathrm{ii}}}+\sqrt{{\left(x+c\right)}^2+y^2}& =\hfill & 2a.\hfill \end{array}$$

Subtract the second radical from both sides and square both sides:

$$\begin{array}{ccc}\hfill \sqrt{{\left(x-c\right)}^2+y^2}& =\hfill & 2a-\sqrt{{\left(\mathrm{ten}+c\right)}^{\mathrm{ii}}+y^{\mathrm{ii}}}\hfill \\ \hfill {\left(\mathrm{10}-c\right)}^2+y^2& =\hfill & 4a^{\mathrm{two}}-4a\sqrt{{\left(x+c\right)}^2+y^2}+{\left(x+c\right)}^2+y^2\hfill \\ \hfill x^2-2cx+c^{\mathrm{ii}}+y^2& =\hfill & 4a^2-\mathrm{iv}a\sqrt{{\left(\mathrm{ten}+c\right)}^2+y^2}+x^{\mathrm{ii}}+2cx+c^2+y^{\mathrm{two}}\hfill \\ \hfill \text{−}2cx& =\hfill & 4a^2-4a\sqrt{{\left(x+c\right)}^{\mathrm{ii}}+y^2}+2cx.\hfill \end{array}$$

Now isolate the radical on the right-mitt side and foursquare again:

$$\begin{array}{ccc}\hfill -\mathrm{ii}cx& =\hfill & \mathrm{four}{a}^{\mathrm{ii}}-4a\sqrt{{\left(x+c\right)}^2+y^2}+\mathrm{ii}c\mathrm{ten}\hfill \\ \hfill 4a\sqrt{{\left(x+c\right)}^2+y^2}& =\hfill & 4a^2+\mathrm{four}cx\hfill \\ \hfill \sqrt{{\left(\mathrm{ten}+c\right)}^{\mathrm{ii}}+y^2}& =\hfill & a+\frac{c\mathrm{10}}{a}\hfill \\ \hfill {\left(x+c\right)}^2+y^2& =\hfill & a^2+2cx+\frac{c^2{x}^2}{a^2}\hfill \\ \hfill x^2+2cx+c^{\mathrm{ii}}+y^2& =\hfill & a^2+2cx+\frac{c^2{x}^2}{a^{\mathrm{ii}}}\hfill \\ \hfill {\mathrm{ten}}^{\mathrm{two}}+c^{\mathrm{ii}}+y^2& =\hfill & a^{\mathrm{two}}+\frac{c^2{x}^2}{a^2}.\hfill \end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the right-hand side:

$$\begin{array}{}\\ \hfill x^{\mathrm{ii}}-\frac{c^2{x}^2}{a^2}+y^2& =\hfill & a^{\mathrm{ii}}-c^2\hfill \\ \hfill \frac{\left(a^2-c^2\right){\mathrm{ten}}^2}{a^2}+y^2& =\hfill & a^2-c^2.\hfill \end{array}$$

Divide both sides by $a^2-c^{\mathrm{two}}.$ This gives the equation

$$\frac{x^2}{a^2}+\frac{y^2}{a^{\mathrm{two}}-c^2}=\mathrm{ane}.$$

If we refer back to Figure 7.48, so the length of each of the two green line segments is equal to a. This is true because the sum of the distances from the point Q to the foci $F\text{and}{F}^{\prime }$ is equal to 2a, and the lengths of these 2 line segments are equal. This line segment forms a correct triangle with hypotenuse length a and leg lengths b and c. From the Pythagorean theorem, $a^2=b^2=c^2$ and $b^2+a^{\mathrm{two}}-c^2.$ Therefore the equation of the ellipse becomes

$$\frac{x^{\mathrm{ii}}}{a^{\mathrm{ii}}}+\frac{y^{\mathrm{two}}}{b^2}=\mathrm{i}.$$

Finally, if the eye of the ellipse is moved from the origin to a bespeak $\left(h,\mathrm{one\; thousand}\right),$ nosotros accept the post-obit standard form of an ellipse.

Theorem vii.nine

Equation of an Ellipse in Standard Form

Consider the ellipse with center $\left(h,k\right),$ a horizontal major axis with length twoa, and a vertical modest centrality with length twob. Then the equation of this ellipse in standard course is

$$\frac{{\left(\mathrm{ten}-h\right)}^2}{a^{\mathrm{two}}}+\frac{{\left(y-k\right)}^2}{b^2}=1$$

(7.12)

and the foci are located at $\left(h\pm c,m\right),$ where $c^{\mathrm{two}}=a^{\mathrm{two}}-b^2.$ The equations of the directrices are $\mathrm{10}=h\pm \frac{a^{\mathrm{two}}}{c}.$

If the major axis is vertical, so the equation of the ellipse becomes

$$\frac{{\left(x-h\right)}^2}{b^{\mathrm{two}}}+\frac{{\left(y-\mathrm{thousand}\right)}^2}{a^2}=1$$

(7.thirteen)

and the foci are located at $\left(h,k\pm c\right),$ where $c^{\mathrm{ii}}=a^2-b^2.$ The equations of the directrices in this case are $y=k\pm \frac{a^2}{c}.$

If the major axis is horizontal, so the ellipse is called horizontal, and if the major axis is vertical, then the ellipse is chosen vertical. The equation of an ellipse is in general form if it is in the grade $A{x}^2+B{y}^2+C\mathrm{10}+Dy+\mathrm{East}=0,$ where A and B are either both positive or both negative. To convert the equation from full general to standard course, use the method of completing the square.

Case vii.20

Finding the Standard Course of an Ellipse

Put the equation $\mathrm{ix}{x}^{\mathrm{ii}}+\mathrm{four}{y}^{\mathrm{two}}-36x+24y+36=0$ into standard form and graph the resulting ellipse.

Checkpoint 7.19

Put the equation $9x^2+\mathrm{sixteen}{y}^2+18x-64y-71=0$ into standard grade and graph the resulting ellipse.

According to Kepler's first police of planetary move, the orbit of a planet effectually the Lord's day is an ellipse with the Sun at one of the foci equally shown in Figure 7.50(a). Because Globe'south orbit is an ellipse, the distance from the Sun varies throughout the year. A commonly held misconception is that World is closer to the Sun in the summer. In fact, in summertime for the northern hemisphere, Earth is further from the Sun than during winter. The departure in flavour is caused by the tilt of World'due south axis in the orbital plane. Comets that orbit the Sun, such as Halley'due south Comet, also take elliptical orbits, every bit do moons orbiting the planets and satellites orbiting Globe.

Ellipses also have interesting reflective properties: A calorie-free ray emanating from one focus passes through the other focus later mirror reflection in the ellipse. The same thing occurs with a audio wave also. The National Statuary Hall in the U.S. Capitol in Washington, DC, is a famous room in an elliptical shape as shown in Figure vii.50(b). This hall served as the coming together place for the U.S. House of Representatives for nearly 50 years. The location of the two foci of this semi-elliptical room are clearly identified by marks on the floor, and even if the room is full of visitors, when two people stand on these spots and speak to each other, they can hear each other much more clearly than they tin hear someone standing shut by. Fable has information technology that John Quincy Adams had his desk located on 1 of the foci and was able to eavesdrop on everyone else in the House without ever needing to stand. Although this makes a good story, it is unlikely to be true, because the original ceiling produced so many echoes that the entire room had to exist hung with carpets to dampen the noise. The ceiling was rebuilt in 1902 and only then did the now-famous whispering issue emerge. Another famous whispering gallery—the site of many marriage proposals—is in Grand Primal Station in New York City.

Figure seven.fifty (a) Earth'due south orbit effectually the Sun is an ellipse with the Sun at i focus. (b) Bronze Hall in the U.S. Capitol is a whispering gallery with an elliptical cross section.

Hyperbolas

A hyperbola can also be defined in terms of distances. In the instance of a hyperbola, there are two foci and two directrices. Hyperbolas also have 2 asymptotes.

Definition

A hyperbola is the set up of all points where the difference between their distances from two fixed points (the foci) is constant.

A graph of a typical hyperbola appears every bit follows.

Figure 7.51 A typical hyperbola in which the difference of the distances from whatever betoken on the ellipse to the foci is constant. The transverse axis is also chosen the major axis, and the conjugate axis is also called the modest axis.

The derivation of the equation of a hyperbola in standard class is virtually identical to that of an ellipse. One slight hitch lies in the definition: The difference between ii numbers is always positive. Let P be a point on the hyperbola with coordinates $\left(\mathrm{ten},y\right).$ So the definition of the hyperbola gives $\left|d\left(P,F_1\right)-d\left(P,F_2\right)\right|=\text{abiding}.$ To simplify the derivation, assume that P is on the correct branch of the hyperbola, so the accented value bars drop. If it is on the left branch, then the subtraction is reversed. The vertex of the right branch has coordinates $\left(a,0\right),$ so

$$d\left(P,F_1\right)-d\left(P,F_2\right)=\left(c+a\right)-\left(c-a\right)=2a.$$

This equation is therefore truthful for any point on the hyperbola. Returning to the coordinates $\left(x,y\right)$ for P:

$$\begin{array}{ccc}\hfill d\left(P,F_{\mathrm{one}}\right)-d\left(P,F_2\right)& =\hfill & \mathrm{ii}a\hfill \\ \hfill \sqrt{{\left(x+c\right)}^2+y^{\mathrm{ii}}}-\sqrt{{\left(x-c\right)}^2+y^{\mathrm{ii}}}& =\hfill & 2a.\hfill \end{array}$$

Add the 2d radical from both sides and foursquare both sides:

$$\begin{array}{ccc}\hfill \sqrt{{\left(\mathrm{10}-c\right)}^{\mathrm{ii}}+y^2}& =\hfill & 2a+\sqrt{{\left(x+c\right)}^2+y^2}\hfill \\ \hfill {\left(x-c\right)}^2+y^2& =\hfill & \mathrm{iv}{a}^2+4a\sqrt{{\left(\mathrm{ten}+c\right)}^{\mathrm{two}}+y^{\mathrm{ii}}}+{\left(\mathrm{10}+c\right)}^2+y^2\hfill \\ \hfill x^{\mathrm{ii}}-2cx+c^2+y^2& =\hfill & 4a^2+4a\sqrt{{\left(x+c\right)}^{\mathrm{two}}+y^2}+x^{\mathrm{two}}+\mathrm{two}cx+c^{\mathrm{two}}+y^2\hfill \\ \hfill \text{−}2c\mathrm{ten}& =\hfill & 4a^{\mathrm{ii}}+4a\sqrt{{\left(x+c\right)}^2+y^2}+2cx.\hfill \end{array}$$

Now isolate the radical on the right-manus side and foursquare once again:

$$\begin{array}{ccc}\hfill -2cx& =\hfill & \mathrm{four}{a}^2+\mathrm{four}a\sqrt{{\left(\mathrm{ten}+c\right)}^2+y^{\mathrm{two}}}+2cx\hfill \\ \hfill 4a\sqrt{{\left(x+c\right)}^{\mathrm{two}}+y^{\mathrm{two}}}& =\hfill & \mathrm{-4}{a}^2-\mathrm{iv}cx\hfill \\ \hfill \sqrt{{\left(x+c\right)}^{\mathrm{ii}}+y^{\mathrm{two}}}& =\hfill & \text{−}a-\frac{cx}{a}\hfill \\ \hfill {\left(x+c\right)}^{\mathrm{ii}}+y^{\mathrm{ii}}& =\hfill & a^{\mathrm{ii}}+2cx+\frac{c^2{x}^2}{a^{\mathrm{two}}}\hfill \\ \hfill {\mathrm{10}}^2+2c\mathrm{ten}+c^{\mathrm{two}}+y^{\mathrm{ii}}& =\hfill & a^2+\mathrm{two}c\mathrm{10}+\frac{c^2{x}^2}{a^2}\hfill \\ \hfill x^{\mathrm{two}}+c^2+y^{\mathrm{two}}& =\hfill & a^2+\frac{c^{\mathrm{ii}}{\mathrm{ten}}^2}{a^2}.\hfill \end{array}$$

Isolate the variables on the left-hand side of the equation and the constants on the correct-hand side:

$$\begin{array}{}\\ \hfill x^{\mathrm{ii}}-\frac{c^2{\mathrm{10}}^2}{a^2}+y^{\mathrm{two}}& =\hfill & a^2-c^{\mathrm{ii}}\hfill \\ \hfill \frac{\left(a^2-c^{\mathrm{two}}\right)x^{\mathrm{two}}}{a^2}+y^{\mathrm{ii}}& =\hfill & a^2-c^2.\hfill \end{array}$$

Finally, dissever both sides by $a^2-c^2.$ This gives the equation

$$\frac{x^2}{a^{\mathrm{ii}}}+\frac{y^{\mathrm{ii}}}{a^{\mathrm{ii}}-c^2}=\mathrm{i}.$$

Nosotros at present ascertain b so that $b^{\mathrm{two}}=c^2-a^2.$ This is possible because $c>a.$ Therefore the equation of the ellipse becomes

$$\frac{x^2}{a^{\mathrm{ii}}}-\frac{y^{\mathrm{two}}}{b^2}=\mathrm{one}.$$

Finally, if the center of the hyperbola is moved from the origin to the point $\left(h,\mathrm{thou}\right),$ nosotros accept the following standard form of a hyperbola.

Theorem 7.10

Equation of a Hyperbola in Standard Form

Consider the hyperbola with center $\left(h,k\right),$ a horizontal major axis, and a vertical minor centrality. So the equation of this ellipse is

$$\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=\mathrm{one}$$

(7.14)

and the foci are located at $\left(h\pm c,k\right),$ where $c^{\mathrm{ii}}=a^{\mathrm{two}}+b^{\mathrm{two}}.$ The equations of the asymptotes are given by $y=k\pm \frac{b}{a}\left(x-h\right).$ The equations of the directrices are

$$\mathrm{ten}=k\pm \frac{a^{\mathrm{two}}}{\sqrt{a^{\mathrm{two}}+b^2}}=h\pm \frac{a^2}{c}.$$

If the major axis is vertical, then the equation of the hyperbola becomes

$$\frac{{\left(y-k\right)}^{\mathrm{two}}}{a^2}-\frac{{\left(x-h\right)}^{\mathrm{two}}}{b^{\mathrm{two}}}=1$$

(seven.15)

and the foci are located at $\left(h,m\pm c\right),$ where $c^2=a^2+b^2.$ The equations of the asymptotes are given past $y=k\pm \frac{a}{b}\left(x-h\right).$ The equations of the directrices are

$$y=k\pm \frac{a^2}{\sqrt{a^2+b^2}}=\mathrm{chiliad}\pm \frac{a^2}{c}.$$

If the major axis (transverse axis) is horizontal, then the hyperbola is chosen horizontal, and if the major centrality is vertical then the hyperbola is called vertical. The equation of a hyperbola is in general grade if it is in the form $A{x}^2+B{y}^{\mathrm{ii}}+C\mathrm{10}+Dy+E=0,$ where A and B have opposite signs. In order to convert the equation from full general to standard form, use the method of completing the foursquare.

Instance 7.21

Finding the Standard Grade of a Hyperbola

Put the equation $\mathrm{ix}{\mathrm{10}}^2-16y^{\mathrm{two}}+36\mathrm{ten}+32y-124=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Checkpoint 7.20

Put the equation $4y^2-9x^2+16y+18\mathrm{ten}-29=0$ into standard form and graph the resulting hyperbola. What are the equations of the asymptotes?

Hyperbolas also have interesting reflective properties. A ray directed toward one focus of a hyperbola is reflected by a hyperbolic mirror toward the other focus. This concept is illustrated in the following figure.

Effigy 7.53 A hyperbolic mirror used to collect light from distant stars.

This property of the hyperbola has important applications. Information technology is used in radio direction finding (since the departure in signals from two towers is constant along hyperbolas), and in the construction of mirrors within telescopes (to reflect calorie-free coming from the parabolic mirror to the eyepiece). Another interesting fact about hyperbolas is that for a comet entering the solar system, if the speed is peachy plenty to escape the Dominicus'south gravitational pull, then the path that the comet takes as it passes through the solar arrangement is hyperbolic.

Eccentricity and Directrix

An alternative fashion to describe a conic section involves the directrices, the foci, and a new property chosen eccentricity. Nosotros will meet that the value of the eccentricity of a conic section tin can uniquely ascertain that conic.

Definition

The eccentricity e of a conic section is divers to be the distance from any bespeak on the conic section to its focus, divided by the perpendicular distance from that point to the nearest directrix. This value is constant for any conic department, and can define the conic section as well:

1. If $e=1,$ the conic is a parabola.
2. If $e<1,$ it is an ellipse.
3. If $\mathrm{eastward}>1,$ it is a hyperbola.

The eccentricity of a circumvolve is null. The directrix of a conic section is the line that, together with the betoken known equally the focus, serves to define a conic section. Hyperbolas and noncircular ellipses have ii foci and two associated directrices. Parabolas accept one focus and one directrix.

The iii conic sections with their directrices appear in the following figure.

Figure vii.54 The three conic sections with their foci and directrices.

Recall from the definition of a parabola that the distance from any point on the parabola to the focus is equal to the distance from that same point to the directrix. Therefore, by definition, the eccentricity of a parabola must be 1. The equations of the directrices of a horizontal ellipse are $\mathrm{ten}=\text{±}\frac{a^2}{c}.$ The right vertex of the ellipse is located at $\left(a,0\right)$ and the right focus is $\left(c,0\right).$ Therefore the distance from the vertex to the focus is $a-c$ and the altitude from the vertex to the right directrix is $\frac{a^{\mathrm{two}}}{c}-a.$ This gives the eccentricity as

$$e=\frac{a-c}{\frac{a^2}{c}-a}=\frac{c\left(a-c\right)}{a^2-ac}=\frac{c\left(a-c\right)}{a\left(a-c\right)}=\frac{c}{a}.$$

Since $c<a,$ this footstep proves that the eccentricity of an ellipse is less than ane. The directrices of a horizontal hyperbola are also located at $x=\text{±}\frac{a^{\mathrm{two}}}{c},$ and a similar adding shows that the eccentricity of a hyperbola is besides $e=\frac{c}{a}.$ However in this case we take $c>a,$ so the eccentricity of a hyperbola is greater than i.

Example 7.22

Determining Eccentricity of a Conic Section

Make up one's mind the eccentricity of the ellipse described past the equation

$$\frac{{\left(x-\mathrm{three}\right)}^{\mathrm{ii}}}{16}+\frac{{\left(y+2\right)}^{\mathrm{two}}}{25}=1.$$

Checkpoint vii.21

Determine the eccentricity of the hyperbola described by the equation

$$\frac{{\left(y-\mathrm{three}\right)}^{\mathrm{ii}}}{49}-\frac{{\left(\mathrm{10}+2\right)}^{\mathrm{ii}}}{25}=1.$$

Polar Equations of Conic Sections

Sometimes information technology is useful to write or identify the equation of a conic section in polar grade. To do this, we need the concept of the focal parameter. The focal parameter of a conic section p is defined equally the altitude from a focus to the nearest directrix. The post-obit table gives the focal parameters for the unlike types of conics, where a is the length of the semi-major axis (i.e., half the length of the major axis), c is the distance from the origin to the focus, and e is the eccentricity. In the instance of a parabola, a represents the altitude from the vertex to the focus.

Conic	e	p
Ellipse	$0<e<1$	$\frac{a^2-c^2}{c}=\frac{a^2\left(\mathrm{one}-e^2\right)}{\mathrm{eastward}}$
Parabola	$\mathrm{due\; east}=1$	$2a$
Hyperbola	$e>1$	$\frac{c^2-a^2}{c}=\frac{a\left(e^2-1\right)}{e}$

Table seven.ane Eccentricities and Focal Parameters of the Conic Sections

Using the definitions of the focal parameter and eccentricity of the conic section, we tin can derive an equation for any conic section in polar coordinates. In particular, we assume that ane of the foci of a given conic section lies at the pole. Then using the definition of the diverse conic sections in terms of distances, it is possible to prove the following theorem.

Theorem 7.11

Polar Equation of Conic Sections

The polar equation of a conic section with focal parameter p is given by

$$r=\frac{\mathrm{due\; east}p}{\mathrm{ane}\pm e\text{cos}\theta }\text{or}r=\frac{\mathrm{east}p}{1\pm \mathrm{due\; east}\text{sin}\theta }.$$

In the equation on the left, the major centrality of the conic department is horizontal, and in the equation on the right, the major axis is vertical. To work with a conic section written in polar grade, first make the constant term in the denominator equal to 1. This can be done by dividing both the numerator and the denominator of the fraction by the constant that appears in front of the plus or minus in the denominator. Then the coefficient of the sine or cosine in the denominator is the eccentricity. This value identifies the conic. If cosine appears in the denominator, then the conic is horizontal. If sine appears, then the conic is vertical. If both appear and then the axes are rotated. The center of the conic is not necessarily at the origin. The heart is at the origin merely if the conic is a circumvolve (i.e., $e=0).$

Example 7.23

Graphing a Conic Section in Polar Coordinates

Identify and create a graph of the conic section described past the equation

$$r=\frac{3}{\mathrm{ane}+2\text{cos}\theta }.$$

Checkpoint 7.22

Identify and create a graph of the conic section described by the equation

$$r=\frac{4}{\mathrm{ane}-\mathrm{0.eight}\text{sin}\theta }.$$

General Equations of Degree 2

A general equation of degree 2 can be written in the form

$$A{x}^2+Bxy+C{y}^{\mathrm{two}}+Dx+\mathrm{Due\; east}y+F=0.$$

The graph of an equation of this class is a conic section. If $B\ne 0$ then the coordinate axes are rotated. To identify the conic department, we use the discriminant of the conic section $4AC-B^2.$ One of the following cases must be true:

1. $\mathrm{four}AC-B^2>0.$ If and so, the graph is an ellipse.
2. $4AC-B^2=0.$ If then, the graph is a parabola.
3. $4AC-B^{\mathrm{ii}}<0.$ If and then, the graph is a hyperbola.

The simplest case of a second-degree equation involving a cross term is $xy=\mathrm{i}.$ This equation can be solved for y to obtain $y=\frac{\mathrm{ane}}{x}.$ The graph of this role is called a rectangular hyperbola every bit shown.

Effigy 7.56 Graph of the equation $xy=\mathrm{ane};$ The red lines indicate the rotated axes.

The asymptotes of this hyperbola are the x and y coordinate axes. To make up one's mind the angle $\theta$ of rotation of the conic section, we utilize the formula $\text{cot}2\theta =\frac{A-C}{B}.$ In this instance $A=C=0$ and $B=\mathrm{one},$ so $\text{cot}\mathrm{ii}\theta =(0-0)\text{/}1=0$ and $\theta =45\text{°}.$ The method for graphing a conic department with rotated axes involves determining the coefficients of the conic in the rotated coordinate system. The new coefficients are labeled $A^{\prime },B^{\prime },C^{\prime },D^{\prime },{\mathrm{Eastward}}^{\prime },\text{and}{F}^{\prime },$ and are given by the formulas

$$\begin{array}{ccc}\hfill A^{\prime }& =\hfill & A{\text{cos}}^{\mathrm{ii}}\theta +B\text{cos}\theta \text{sin}\theta +C{\text{sin}}^2\theta \hfill \\ \hfill B^{\prime }& =\hfill & 0\hfill \\ \hfill C^{\prime }& =\hfill & A{\text{sin}}^2\theta -B\text{sin}\theta \text{cos}\theta +C{\text{cos}}^{\mathrm{ii}}\theta \hfill \\ \hfill D^{\prime }& =\hfill & D\text{cos}\theta +E\text{sin}\theta \hfill \\ \hfill {\mathrm{East}}^{\prime }& =\hfill & \text{−}D\text{sin}\theta +\mathrm{East}\text{cos}\theta \hfill \\ \hfill F^{\prime }& =\hfill & F.\hfill \end{array}$$

The procedure for graphing a rotated conic is the following:

1. Place the conic section using the discriminant $\mathrm{four}AC-B^{\mathrm{two}}.$
2. Determine $\theta$ using the formula $\text{cot}\mathrm{two}\theta =\frac{A-C}{B}.$
3. Summate $A^{\prime },B^{\prime },C^{\prime },D^{\prime },{\mathrm{East}}^{\prime },\text{and}{F}^{\prime }.$
4. Rewrite the original equation using $A^{\prime },B^{\prime },C^{\prime },D^{\prime },E^{\prime },\text{and}{F}^{\prime }.$
5. Describe a graph using the rotated equation.

Example 7.24

Identifying a Rotated Conic

Identify the conic and calculate the angle of rotation of axes for the bend described by the equation

$$\mathrm{thirteen}{\mathrm{ten}}^{\mathrm{two}}-\mathrm{six}\sqrt{3}xy+7y^2-256=0.$$

Checkpoint vii.23

Identify the conic and calculate the angle of rotation of axes for the curve described by the equation

$$3{\mathrm{10}}^2+5xy-2y^{\mathrm{ii}}-125=0.$$

Section 7.v Exercises

For the following exercises, determine the equation of the parabola using the data given.

255.

Focus $\left(4,0\right)$ and directrix $x=\mathrm{-4}$

256 .

Focus $\left(0,\mathrm{-3}\right)$ and directrix $y=\mathrm{three}$

257.

Focus $\left(0,\mathrm{0.five}\right)$ and directrix $y=\mathrm{-0.v}$

258 .

Focus $\left(\mathrm{two},3\right)$ and directrix $\mathrm{ten}=\mathrm{-two}$

259.

Focus $\left(0,2\right)$ and directrix $y=4$

260 .

Focus $\left(\mathrm{-1},4\right)$ and directrix $x=\mathrm{v}$

261.

Focus $\left(\mathrm{-iii},\mathrm{five}\right)$ and directrix $y=\mathrm{ane}$

262 .

Focus $\left(\frac{\mathrm{five}}{2},\mathrm{-4}\right)$ and directrix $x=\frac{\mathrm{vii}}{2}$

For the following exercises, make up one's mind the equation of the ellipse using the data given.

263.

Endpoints of major axis at $\left(4,0\right),\left(\mathrm{-4},0\right)$ and foci located at $\left(\mathrm{two},0\right),\left(\mathrm{-2},0\right)$

264 .

Endpoints of major axis at $\left(0,\mathrm{five}\right),\left(0,\mathrm{-5}\right)$ and foci located at $\left(0,\mathrm{iii}\right),\left(0,\mathrm{-3}\right)$

265.

Endpoints of pocket-sized axis at $\left(0,\mathrm{two}\right),\left(0,\mathrm{-ii}\right)$ and foci located at $\left(\mathrm{iii},0\right),\left(\mathrm{-3},0\right)$

266 .

Endpoints of major axis at $\left(\mathrm{-3},3\right),\left(\mathrm{vii},3\right)$ and foci located at $\left(\mathrm{-2},3\right),\left(\mathrm{half-dozen},3\right)$

267.

Endpoints of major axis at $\left(\mathrm{-3},5\right),\left(\mathrm{-3},\mathrm{-3}\right)$ and foci located at $\left(\mathrm{-3},3\right),\left(\mathrm{-3},\mathrm{-1}\right)$

268 .

Endpoints of major axis at $\left(0,0\right),\left(0,4\right)$ and foci located at $\left(5,2\right),\left(\mathrm{-5},2\right)$

269.

Foci located at $\left(2,0\right),\left(\mathrm{-ii},0\right)$ and eccentricity of $\frac{1}{2}$

270 .

Foci located at $\left(0,\mathrm{-3}\right),\left(0,3\right)$ and eccentricity of $\frac{\mathrm{iii}}{\mathrm{iv}}$

For the following exercises, make up one's mind the equation of the hyperbola using the information given.

271.

Vertices located at $\left(5,0\right),\left(\mathrm{-5},0\right)$ and foci located at $\left(\mathrm{six},0\right),\left(\mathrm{-6},0\right)$

272 .

Vertices located at $\left(0,2\right),\left(0,\mathrm{-2}\right)$ and foci located at $\left(0,\mathrm{iii}\right),\left(0,\mathrm{-3}\right)$

273.

Endpoints of the conjugate axis located at $\left(0,3\right),\left(0,\mathrm{-three}\right)$ and foci located $\left(\mathrm{iv},0\right),\left(\mathrm{-4},0\right)$

274 .

Vertices located at $\left(0,1\right),\left(6,1\right)$ and focus located at $\left(8,\mathrm{ane}\right)$

275.

Vertices located at $\left(\mathrm{-2},0\right),\left(\mathrm{-2},\mathrm{-4}\right)$ and focus located at $\left(\mathrm{-two},\mathrm{-8}\right)$

276 .

Endpoints of the conjugate axis located at $\left(\mathrm{three},\mathrm{two}\right),\left(\mathrm{three},4\right)$ and focus located at $\left(\mathrm{three},7\right)$

277.

Foci located at $\left(\mathrm{six},\mathrm{-0}\right),\left(6,0\right)$ and eccentricity of iii

278 .

$\left(0,10\right),\left(0,\mathrm{-x}\right)$ and eccentricity of two.5

For the following exercises, consider the following polar equations of conics. Decide the eccentricity and place the conic.

279.

$r=\frac{\mathrm{-one}}{1+\text{cos}\theta }$

280 .

$r=\frac{8}{\mathrm{ii}-\text{sin}\theta }$

281.

$r=\frac{\mathrm{v}}{2+\text{sin}\theta }$

282 .

$r=\frac{5}{\mathrm{-ane}+\mathrm{two}\text{sin}\theta }$

283.

$r=\frac{3}{2-6\text{sin}\theta }$

284 .

$r=\frac{\mathrm{iii}}{\mathrm{-four}+3\text{sin}\theta }$

For the following exercises, find a polar equation of the conic with focus at the origin and eccentricity and directrix as given.

285.

$\text{Directrix:}\mathrm{10}=4;e=\frac{1}{\mathrm{v}}$

286 .

$\text{Directrix:}x=\mathrm{-4};e=5$

287.

$\text{Directrix: y}=2;e=2$

288 .

$\text{Directrix: y}=\mathrm{-2};e=\frac{\mathrm{i}}{2}$

For the following exercises, sketch the graph of each conic.

289.

$r=\frac{1}{1+\text{sin}\theta }$

290 .

$r=\frac{\mathrm{one}}{1-\text{cos}\theta }$

291.

$r=\frac{4}{1+\text{cos}\theta }$

292 .

$r=\frac{10}{5+4\text{sin}\theta }$

293.

$r=\frac{15}{\mathrm{three}-\mathrm{two}\text{cos}\theta }$

294 .

$r=\frac{32}{3+\mathrm{five}\text{sin}\theta }$

295.

$r(\mathrm{two}+\text{sin}\theta )=4$

296 .

$r=\frac{3}{\mathrm{ii}+6\text{sin}\theta }$

297.

$r=\frac{\mathrm{three}}{\mathrm{-4}+2\text{sin}\theta }$

298 .

$\begin{array}{c}\frac{{\mathrm{ten}}^2}{9}+\frac{y^2}{\mathrm{four}}=1\hfill \\ \hfill \end{array}$

299.

$\frac{{\mathrm{ten}}^2}{\mathrm{iv}}+\frac{y^2}{16}=1$

300 .

$\mathrm{four}{x}^{\mathrm{ii}}+\mathrm{nine}{y}^2=36$

301.

$25x^2-4y^2=100$

302 .

$\frac{x^{\mathrm{ii}}}{16}-\frac{y^2}{9}=\mathrm{one}$

303.

${\mathrm{10}}^2=12y$

304 .

$y^{\mathrm{two}}=20\mathrm{ten}$

305.

$12\mathrm{10}=5y^{\mathrm{ii}}$

For the following equations, make up one's mind which of the conic sections is described.

307.

$x^2+4\mathrm{10}y-\mathrm{two}{y}^{\mathrm{two}}-\mathrm{half-dozen}=0$

308 .

$x^2+2\sqrt{3}xy+3y^{\mathrm{two}}-\mathrm{half\; dozen}=0$

309.

${\mathrm{ten}}^2-xy+y^2-2=0$

310 .

$34x^{\mathrm{ii}}-24\mathrm{10}y+41y^2-25=0$

311.

$52{\mathrm{10}}^{\mathrm{ii}}-72\mathrm{ten}y+73y^2+\mathrm{forty}x+30y-75=0$

312 .

The mirror in an automobile headlight has a parabolic cross section, with the lightbulb at the focus. On a schematic, the equation of the parabola is given every bit $x^2=4y.$ At what coordinates should you place the lightbulb?

313.

A satellite dish is shaped like a paraboloid of revolution. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver exist placed?

314 .

Consider the satellite dish of the preceding trouble. If the dish is 8 anxiety across at the opening and 2 feet deep, where should we place the receiver?

315.

A searchlight is shaped similar a paraboloid of revolution. A calorie-free source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 3 feet across, detect the depth.

316 .

Whispering galleries are rooms designed with elliptical ceilings. A person standing at one focus can whisper and exist heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet and the foci are located xxx feet from the center, find the pinnacle of the ceiling at the center.

317.

A person is standing eight feet from the nearest wall in a whispering gallery. If that person is at one focus and the other focus is 80 feet away, what is the length and the peak at the center of the gallery?

For the post-obit exercises, determine the polar equation form of the orbit given the length of the major centrality and eccentricity for the orbits of the comets or planets. Distance is given in astronomical units (AU).

318 .

Halley's Comet: length of major centrality = 35.88, eccentricity = 0.967

319.

Hale-Bopp Comet: length of major axis = 525.91, eccentricity = 0.995

320 .

Mars: length of major axis = three.049, eccentricity = 0.0934

321.

Jupiter: length of major centrality = x.408, eccentricity = 0.0484

Eccentricity Of A Conic Section,

Source: https://openstax.org/books/calculus-volume-2/pages/7-5-conic-sections

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