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10.6: Conic Sections in Polar Coordinates

  • Page ID
    114102
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    Learning Objectives

    In this section, you will:

    • Identify a conic in polar form.
    • Graph the polar equations of conics.
    • Define conics in terms of a focus and a directrix.
    1136cb526580abf04edf928250ffbf169f61cf1d
    Figure 1 Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr)

    Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits.

    In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system.

    Identifying a Conic in Polar Form

    Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola x=2+ y 2 x=2+ y 2 shown in Figure 2.

    A horizontal parabola, labeled x = 2 + y squared, opening to the right is shown. The Focus is labeled Focus @ pole and is on the horizontal Polar Axis. The vertical Directrix is shown. A point on the upper side of the parabola is labeled P times (r, theta) and two lines of equal length r are drawn from it, one to the Focus and the other to the Directrix and perpendicular to it. The line to the Focus makes an angle theta with the Polar Axis.
    Figure 2

    In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus P(r,θ) P(r,θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis.

    If F F is a fixed point, the focus, and D D is a fixed line, the directrix, then we can let e e be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points P P such that e= PF PD e= PF PD is a conic. In other words, we can define a conic as the set of all points P P with the property that the ratio of the distance from P P to F F to the distance from P P to D D is equal to the constant e. e.

    For a conic with eccentricity e, e,

    • if 0e<1, 0e<1, the conic is an ellipse
    • if e=1, e=1, the conic is a parabola
    • if e>1, e>1, the conic is an hyperbola

    With this definition, we may now define a conic in terms of the directrix, x=±p, x=±p, the eccentricity e, e, and the angle θ. θ. Thus, each conic may be written as a polar equation, an equation written in terms of r r and θ. θ.

    The Polar Equation for a Conic

    For a conic with a focus at the origin, if the directrix is x=±p, x=±p, where p p is a positive real number, and the eccentricity is a positive real number e, e, the conic has a polar equation

    r= ep 1±ecosθ r= ep 1±ecosθ

    For a conic with a focus at the origin, if the directrix is y=±p, y=±p, where p p is a positive real number, and the eccentricity is a positive real number e, e, the conic has a polar equation

    r= ep 1±esinθ r= ep 1±esinθ

    How To

    Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity.

    1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form.
    2. Identify the eccentricity e e as the coefficient of the trigonometric function in the denominator.
    3. Compare e e with 1 to determine the shape of the conic.
    4. Determine the directrix as x=p x=p if cosine is in the denominator and y=p y=p if sine is in the denominator. Set ep ep equal to the numerator in standard form to solve for x x or y. y.

    Example 1

    Identifying a Conic Given the Polar Form

    For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity.

    1. r= 6 3+2sinθ r= 6 3+2sinθ
    2. r= 12 4+5cosθ r= 12 4+5cosθ
    3. r= 7 22sinθ r= 7 22sinθ
    Answer

    For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, 1 c , 1 c , where c c is that constant.

    1. r= 6 3+2sinθ ( 1 3 ) ( 1 3 ) = 6( 1 3 ) 3( 1 3 )+2( 1 3 )sinθ = 2 1+ 2 3 sinθ r= 6 3+2sinθ ( 1 3 ) ( 1 3 ) = 6( 1 3 ) 3( 1 3 )+2( 1 3 )sinθ = 2 1+ 2 3 sinθ

      2=ep 2= 2 3 p ( 3 2 )2=( 3 2 ) 2 3 p 3=p 2=ep 2= 2 3 p ( 3 2 )2=( 3 2 ) 2 3 p 3=p

      Since e<1, e<1, the conic is an ellipse. The eccentricity is e= 2 3 e= 2 3 and the directrix is y=3. y=3.

    2. r= 12 4+5cosθ ( 1 4 ) ( 1 4 ) r= 12( 1 4 ) 4( 1 4 )+5( 1 4 )cosθ r= 3 1+ 5 4 cosθ r= 12 4+5cosθ ( 1 4 ) ( 1 4 ) r= 12( 1 4 ) 4( 1 4 )+5( 1 4 )cosθ r= 3 1+ 5 4 cosθ

      3=ep 3= 5 4 p ( 4 5 )3=( 4 5 ) 5 4 p 12 5 =p 3=ep 3= 5 4 p ( 4 5 )3=( 4 5 ) 5 4 p 12 5 =p

      Since e>1, e>1, the conic is a hyperbola. The eccentricity is e= 5 4 e= 5 4 and the directrix is x= 12 5 =2.4. x= 12 5 =2.4.

    3. r= 7 22sinθ ( 1 2 ) ( 1 2 ) r= 7( 1 2 ) 2( 1 2 )2( 1 2 )sinθ r= 7 2 1sinθ r= 7 22sinθ ( 1 2 ) ( 1 2 ) r= 7( 1 2 ) 2( 1 2 )2( 1 2 )sinθ r= 7 2 1sinθ

      7 2 =ep 7 2 =( 1 )p 7 2 =p 7 2 =ep 7 2 =( 1 )p 7 2 =p

      Because e=1, e=1, the conic is a parabola. The eccentricity is e=1 e=1 and the directrix is y= 7 2 =−3.5. y= 7 2 =−3.5.

    Try It #1

    Identify the conic with focus at the origin, the directrix, and the eccentricity for r= 2 3cosθ . r= 2 3cosθ .

    Graphing the Polar Equations of Conics

    When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine e e and, therefore, the shape of the curve. The next step is to substitute values for θ θ and solve for r r to plot a few key points. Setting θ θ equal to 0, π 2 ,π, 0, π 2 ,π, and 3π 2 3π 2 provides the vertices so we can create a rough sketch of the graph.

    Example 2

    Graphing a Parabola in Polar Form

    Graph r= 5 3+3cosθ . r= 5 3+3cosθ .

    Answer

    First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is 1 3 . 1 3 .

    r= 5 3+3cosθ = 5( 1 3 ) 3( 1 3 )+3( 1 3 )cosθ r= 5 3 1+cosθ r= 5 3+3cosθ = 5( 1 3 ) 3( 1 3 )+3( 1 3 )cosθ r= 5 3 1+cosθ

    Because e=1, e=1, we will graph a parabola with a focus at the origin. The function has a cosθ, cosθ, and there is an addition sign in the denominator, so the directrix is x=p. x=p.

    5 3 =ep 5 3 =(1)p 5 3 =p 5 3 =ep 5 3 =(1)p 5 3 =p

    The directrix is x= 5 3 . x= 5 3 .

    Plotting a few key points as in Table 1 will enable us to see the vertices. See Figure 3.

    A B C D
    θ θ 0 0 π 2 π 2 π π 3π 2 3π 2
    r= 5 3+3cosθ r= 5 3+3cosθ 5 6 0.83 5 6 0.83 5 3 1.67 5 3 1.67 undefined 5 3 1.67 5 3 1.67
    Table 1
    A horizontal parabola opening left is shown in a polar coordinate system. The Focus is at the Pole. The Directrix, the vertical line x = 5/3, is shown. The Vertex is labeled A. The points where the parabola intersects the vertical axis through the Pole are labeled: the upper point is B, the lower point is D. The Polar Axis tick marks are labeled 2, 3, 4, 5.
    Figure 3

    Analysis

    We can check our result with a graphing utility. See Figure 4.

    A horizontal parabola opening left is shown in a polar coordinate system. The Vertex is on the Polar Axis at r = 1. The Polar Axis tick marks are labeled 2, 3, 4, 5.
    Figure 4

    Example 3

    Graphing a Hyperbola in Polar Form

    Graph r= 8 23sinθ . r= 8 23sinθ .

    Answer

    First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 1 2 . 1 2 .

    r= 8 23sinθ = 8( 1 2 ) 2( 1 2 )3( 1 2 )sinθ r= 4 1 3 2 sinθ r= 8 23sinθ = 8( 1 2 ) 2( 1 2 )3( 1 2 )sinθ r= 4 1 3 2 sinθ

    Because e= 3 2 ,e>1, e= 3 2 ,e>1, so we will graph a hyperbola with a focus at the origin. The function has a sinθ sinθ term and there is a subtraction sign in the denominator, so the directrix is y=p. y=p.

    4=ep 4=( 3 2 )p 4( 2 3 )=p 8 3 =p 4=ep 4=( 3 2 )p 4( 2 3 )=p 8 3 =p

    The directrix is y= 8 3 . y= 8 3 .

    Plotting a few key points as in Table 2 will enable us to see the vertices. See Figure 5.

    A B C D
    θ θ 0 0 π 2 π 2 π π 3π 2 3π 2
    r= 8 23sinθ r= 8 23sinθ 4 4 8 8 4 4 8 5 =1.6 8 5 =1.6
    Table 2
    A vertical hyperbola is shown in a polar coordinate system, centered below the Pole. The Vertices are on the vertical axis through the Pole. The upper Vertex is labeled D and the lower Vertex is labeled B. The points where the upper branch of the hyperbola intersect the Polar Axis and its horizontal extension are labeled A and C respectively. The Polar Axis tick marks are labeled 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
    Figure 5

    Example 4

    Graphing an Ellipse in Polar Form

    Graph r= 10 54cosθ . r= 10 54cosθ .

    Answer

    First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 1 5 . 1 5 .

    r= 10 54cosθ = 10( 1 5 ) 5( 1 5 )4( 1 5 )cosθ r= 2 1 4 5 cosθ r= 10 54cosθ = 10( 1 5 ) 5( 1 5 )4( 1 5 )cosθ r= 2 1 4 5 cosθ

    Because e= 4 5 ,e<1, e= 4 5 ,e<1, so we will graph an ellipse with a focus at the origin. The function has a cosθ, cosθ, and there is a subtraction sign in the denominator, so the directrix is x=p. x=p.

    2=ep 2=( 4 5 )p 2( 5 4 )=p 5 2 =p 2=ep 2=( 4 5 )p 2( 5 4 )=p 5 2 =p

    The directrix is x= 5 2 . x= 5 2 .

    Plotting a few key points as in Table 3 will enable us to see the vertices. See Figure 6.

    A B C D
    θ θ 0 0 π 2 π 2 π π 3π 2 3π 2
    r= 10 54cosθ r= 10 54cosθ 10 10 2 2 10 9 1.1 10 9 1.1 2 2
    Table 3
    A horizontal ellipse is shown in a polar coordinate system, centered on the Polar Axis to the right of the Pole. The Vertices are on the Polar Axis. The right Vertex is labeled A and the left Vertex is labeled C and is to the left of the Pole.  Point A is on the Polar Axis at r = 10. The Polar Axis tick marks are labeled 2, 4, 6, 8, 10, 12. The upper and lower points where the ellipse intersects the vertical axis through the Pole are labeled B and D respectively. The Directrix, the vertical line x = negative 5/2, is shown.
    Figure 6

    Analysis

    We can check our result using a graphing utility. See Figure 7.

    810c054eafe0e3d4f54ddd306dc735053119214b
    Figure 7 r= 10 54cosθ r= 10 54cosθ graphed on a viewing window of [ –3,12,1 ] [ –3,12,1 ] by [4,4,1],θmin =0 [4,4,1],θmin =0 and θmax =2π. θmax =2π.

    Try It #2

    Graph r= 2 4cosθ . r= 2 4cosθ .

    Defining Conics in Terms of a Focus and a Directrix

    So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.

    How To

    Given the focus, eccentricity, and directrix of a conic, determine the polar equation.

    1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y, y, we use the general polar form in terms of sine. If the directrix is given in terms of x, x, we use the general polar form in terms of cosine.
    2. Determine the sign in the denominator. If p<0, p<0, use subtraction. If p>0, p>0, use addition.
    3. Write the coefficient of the trigonometric function as the given eccentricity.
    4. Write the absolute value of p p in the numerator, and simplify the equation.

    Example 5

    Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix

    Find the polar form of the conic given a focus at the origin, e=3 e=3 and directrix y=2. y=2.

    Answer

    The directrix is y=p, y=p, so we know the trigonometric function in the denominator is sine.

    Because y=−2,–2<0, y=−2,–2<0, so we know there is a subtraction sign in the denominator. We use the standard form of

    r= ep 1esinθ r= ep 1esinθ

    and e=3 e=3 and | −2 |=2=p. | −2 |=2=p.

    Therefore,

    r= (3)(2) 13sinθ r= 6 13sinθ r= (3)(2) 13sinθ r= 6 13sinθ

    Example 6

    Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix

    Find the polar form of a conic given a focus at the origin, e= 3 5 , e= 3 5 , and directrix x=4. x=4.

    Answer

    Because the directrix is x=p, x=p, we know the function in the denominator is cosine. Because x=4,4>0, x=4,4>0, so we know there is an addition sign in the denominator. We use the standard form of

    r= ep 1+ecosθ r= ep 1+ecosθ

    and e= 3 5 e= 3 5 and | 4 |=4=p. | 4 |=4=p.

    Therefore,

    r= ( 3 5 )(4) 1+ 3 5 cosθ r= 12 5 1+ 3 5 cosθ r= 12 5 1( 5 5 )+ 3 5 cosθ r= 12 5 5 5 + 3 5 cosθ r= 12 5 5 5+3cosθ r= 12 5+3cosθ r= ( 3 5 )(4) 1+ 3 5 cosθ r= 12 5 1+ 3 5 cosθ r= 12 5 1( 5 5 )+ 3 5 cosθ r= 12 5 5 5 + 3 5 cosθ r= 12 5 5 5+3cosθ r= 12 5+3cosθ

    Try It #3

    Find the polar form of the conic given a focus at the origin, e=1, e=1, and directrix x=−1. x=−1.

    Example 7

    Converting a Conic in Polar Form to Rectangular Form

    Convert the conic r= 1 55sinθ r= 1 55sinθ to rectangular form.

    Answer

    We will rearrange the formula to use the identities r= x 2 + y 2 ,x=rcosθ,and y=rsinθ. r= x 2 + y 2 ,x=rcosθ,and y=rsinθ.

    r= 1 55sinθ r(55sinθ)= 1 55sinθ (55sinθ) Eliminate the fraction. 5r5rsinθ=1 Distribute. 5r=1+5rsinθ Isolate 5r. 25 r 2 = (1+5rsinθ) 2 Square both sides. 25( x 2 + y 2 )= (1+5y) 2 Substitute r= x 2 + y 2 and y=rsinθ. 25 x 2 +25 y 2 =1+10y+25 y 2 Distribute and use FOIL. 25 x 2 10y=1 Rearrange terms and set equal to 1. r= 1 55sinθ r(55sinθ)= 1 55sinθ (55sinθ) Eliminate the fraction. 5r5rsinθ=1 Distribute. 5r=1+5rsinθ Isolate 5r. 25 r 2 = (1+5rsinθ) 2 Square both sides. 25( x 2 + y 2 )= (1+5y) 2 Substitute r= x 2 + y 2 and y=rsinθ. 25 x 2 +25 y 2 =1+10y+25 y 2 Distribute and use FOIL. 25 x 2 10y=1 Rearrange terms and set equal to 1.

    Try It #4

    Convert the conic r= 2 1+2cosθ r= 2 1+2cosθ to rectangular form.

    Media

    Access these online resources for additional instruction and practice with conics in polar coordinates.

    10.5 Section Exercises

    Verbal

    1.

    Explain how eccentricity determines which conic section is given.

    2.

    If a conic section is written as a polar equation, what must be true of the denominator?

    3.

    If a conic section is written as a polar equation, and the denominator involves sinθ, sinθ, what conclusion can be drawn about the directrix?

    4.

    If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph?

    5.

    What do we know about the focus/foci of a conic section if it is written as a polar equation?

    Algebraic

    For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity.

    6.

    r= 6 12cosθ r= 6 12cosθ

    7.

    r= 3 44sinθ r= 3 44sinθ

    8.

    r= 8 43cosθ r= 8 43cosθ

    9.

    r= 5 1+2sinθ r= 5 1+2sinθ

    10.

    r= 16 4+3cosθ r= 16 4+3cosθ

    11.

    r= 3 10+10cosθ r= 3 10+10cosθ

    12.

    r= 2 1cosθ r= 2 1cosθ

    13.

    r= 4 7+2cosθ r= 4 7+2cosθ

    14.

    r(1cosθ)=3 r(1cosθ)=3

    15.

    r(3+5sinθ)=11 r(3+5sinθ)=11

    16.

    r(45sinθ)=1 r(45sinθ)=1

    17.

    r(7+8cosθ)=7 r(7+8cosθ)=7

    For the following exercises, convert the polar equation of a conic section to a rectangular equation.

    18.

    r= 4 1+3sinθ r= 4 1+3sinθ

    19.

    r= 2 53sinθ r= 2 53sinθ

    20.

    r= 8 32cosθ r= 8 32cosθ

    21.

    r= 3 2+5cosθ r= 3 2+5cosθ

    22.

    r= 4 2+2sinθ r= 4 2+2sinθ

    23.

    r= 3 88cosθ r= 3 88cosθ

    24.

    r= 2 6+7cosθ r= 2 6+7cosθ

    25.

    r= 5 511sinθ r= 5 511sinθ

    26.

    r(5+2cosθ)=6 r(5+2cosθ)=6

    27.

    r(2cosθ)=1 r(2cosθ)=1

    28.

    r(2.52.5sinθ)=5 r(2.52.5sinθ)=5

    29.

    r= 6secθ 2+3secθ r= 6secθ 2+3secθ

    30.

    r= 6cscθ 3+2cscθ r= 6cscθ 3+2cscθ

    For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.

    31.

    r= 5 2+cosθ r= 5 2+cosθ

    32.

    r= 2 3+3sinθ r= 2 3+3sinθ

    33.

    r= 10 54sinθ r= 10 54sinθ

    34.

    r= 3 1+2cosθ r= 3 1+2cosθ

    35.

    r= 8 45cosθ r= 8 45cosθ

    36.

    r= 3 44cosθ r= 3 44cosθ

    37.

    r= 2 1sinθ r= 2 1sinθ

    38.

    r= 6 3+2sinθ r= 6 3+2sinθ

    39.

    r(1+cosθ)=5 r(1+cosθ)=5

    40.

    r(34sinθ)=9 r(34sinθ)=9

    41.

    r(32sinθ)=6 r(32sinθ)=6

    42.

    r(64cosθ)=5 r(64cosθ)=5

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

    43.

    Directrix: x=4;e= 1 5 x=4;e= 1 5

    44.

    Directrix: x=4;e=5 x=4;e=5

    45.

    Directrix: y=2;e=2 y=2;e=2

    46.

    Directrix: y=2;e= 1 2 y=2;e= 1 2

    47.

    Directrix: x=1;e=1 x=1;e=1

    48.

    Directrix: x=1;e=1 x=1;e=1

    49.

    Directrix: x= 1 4 ;e= 7 2 x= 1 4 ;e= 7 2

    50.

    Directrix: y= 2 5 ;e= 7 2 y= 2 5 ;e= 7 2

    51.

    Directrix: y=4;e= 3 2 y=4;e= 3 2

    52.

    Directrix: x=−2;e= 8 3 x=−2;e= 8 3

    53.

    Directrix: x=−5;e= 3 4 x=−5;e= 3 4

    54.

    Directrix: y=2;e=2.5 y=2;e=2.5

    55.

    Directrix: x=−3;e= 1 3 x=−3;e= 1 3

    Extensions

    Recall from Rotation of Axes that equations of conics with an xy xy term have rotated graphs. For the following exercises, express each equation in polar form with r r as a function of θ. θ.

    56.

    xy=2 xy=2

    57.

    x 2 +xy+ y 2 =4 x 2 +xy+ y 2 =4

    58.

    2 x 2 +4xy+2 y 2 =9 2 x 2 +4xy+2 y 2 =9

    59.

    16 x 2 +24xy+9 y 2 =4 16 x 2 +24xy+9 y 2 =4

    60.

    2xy+y=1 2xy+y=1


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