Trigonometry/ProOneSoftware-Trigonometry-Win31.iso/trig/chapter7.5r

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 357
à 7.5ïConic Sections in Polar Form

äïPlease give the eccentricity, the equation of the
êêdirectrix, and the name of each of the following conics.

âëGive the eccentricity, directrix, and name of the conic,
êêêër = 3/(2 - 2∙cos Θ).

êë3/2
#ër = ─────────,ëε = 1,ëx = -(3/2)/ε = -3/2,ëparabola
êï1 - cos Θ
éSïIn the previous sections of this chapter, we have looked at
equations of the four conic sections in rectangular form.ïIn the rec-
tangular coordinate system, it is convenient to locate the origin at the
center of the conic section.ïThis location of the origin at the center
generally produces the simplest equation of the conic.ïIn the polar
coordinate system, however, it is more convenient to locate the origin
or pole at a focus of the conic section rather than the center.ïThis
location of the pole will allow us to represent all four conics with
just one polar equation form.
è A conic is defined to be the set of all points in the plane such
that the distance from ç points to a fixed point, called the focus,
divided by the distance from ç points to a fixed line, called the
directrix, is a constant ratio.ïThis constant ratio is called the
eccentricity of the conic.
êêêè In the figure, the distance from the pole to
êêêèthe point P is r.ïThe distance from P to the
êêêèdirectrix is p + r∙cos Θ.ïThe eccentricity,ε,
êêêèis the ratio of ç two distances. ε =
êêêèr/(p + r∙cos Θ).ïSolving this equation for r
êêêègives, r = ε∙p/(1 - ε∙cos Θ).ïWhen the equation
êêêèof the conic is in this form, the coefficient of
@fig7501.bmp,10,145
êêêècos Θ is the eccentricity.ïIf the eccentricity
êêêèis less than 1, the conic is an ellipse.ïIf the
êêêèeccentricity is equal to 1, the conic is a para-
êêêèbola.ïFinally, if the eccentricity is greater
êêêèthan one, the conic is a hyperbola.ïSince the
êêêènumerator of r = ε∙p/(1 - ε∙cos Θ) involves the
êêêèknown value of ε, you can solve for p.ïThis is
êêêèthe distance to the directrix from the origin.
êêêèThus the equation of the directrix is x = -p.
êêêêIn a slightly different orientation with the
êêêèconic opening to the left, the equation becomes
êêêèr = ε∙p/(1 + ε∙cos Θ).ïThe positive sign indi-
êêêècates that the conic opens to the left, and the
êêêèequation of the directrix is x = p.

 1èGive the eccentricity, directrix, and name of the conic,
êêêër = 4/(2 - 2∙cos Θ).

êêêA)ïε = 1,ïx = -2,ïparabola
êêêB)ïε = 2,ïx = -2,ïhyperbola
êêêC)ïε = 1/2,ïx = -4,ïellipse
êêêD)ïå of ç
ü

êè 4êë 2
#èr = ─────────── = ─────────,ïε = 1,ïx = -2/1 = -2,ïparabola
ê2 - 2∙cos Θè1 - cos Θ
Ç A
 2èGive the eccentricity, directrix, and name of the conic,
êêêër = 4/(6 + 3∙cos Θ).

êêêA)ïε = 1,ïx = 4/3,ïparabola
êêêB)ïε = 2,ïx = 3/4,ïhyperbola
êêêC)ïε = 1/2,ïx = 4/3,ïellipse
êêêD)ïå of ç
ü

êè 4êê 2/3êêêï2/3
#èr = ─────────── = ───────────────,ïε = 1/2,ïx = ─── = 4/3,ïellipse
ê6 + 3∙cos Θè1 + (1/2)∙cos Θêêè1/2
Ç C
 3èGive the eccentricity, directrix, and name of the conic,
êêêër = 2/(3 - 4∙cos Θ).

êêêA)ïε = 2/3,ïx = -1/2,ïellipse
êêêB)ïε = 4/3,ïx = -1/2,ïhyperbola
êêêC)ïε = 1,ïx = -1/2,ïparabola
êêêD)ïå of ç
ü

êï2êê 2/3êêê -2/3
# r = ─────────── = ───────────────,ïε = 4/3,ïx = ─── = -1/2, hyperbola
ë3 - 4∙cos Θè1 - (4/3)∙cos Θêêè4/3
Ç B
 4èGive the eccentricity, directrix, and name of the conic,
êêêêër = 2.

êêêA)ïNo graph
êêêB)ïε = 0,ïcircle
êêêC)ïline
êêêD)ïå of ç
ü


 This is the equation of a circle of radius 2.ïIf you use your imagina-
tion, you could say that this conic has two foci that coincide and a
directrix at infinity.ïThus, the distance from the focus to a point
divided by the distance from the point to the directrix is zero.
Ç B
 5èGive the eccentricity, directrix, and name of the conic,
êêêër = 1/(4 - 3∙cos Θ).

êêêA)ïε = 3/4,ïx = -1/3,ïellipse
êêêB)ïε = 1,ïx = 3,ïparabola
êêêC)ïε = 4/3,ïx = -3,ïhyperbola
êêêD)ïå of ç
ü

êï1êê 1/4êêê -1/4
# r = ─────────── = ───────────────,ïε = 3/4,ïx = ─── = -1/3, ellipse
ë4 - 3∙cos Θè1 - (3/4)∙cos Θêêè3/4
Ç A
 6èGive the eccentricity, directrix, and name of the conic,
êêêër = 3/(2 + 2∙cos Θ).

êêêA)ïε = 2/3,ïx = 1/2,ïellipse
êêêB)ïε = 1,ïx = 3/2,ïparabola
êêêC)ïε = 4/3,ïx = 3/4,ïhyperbola
êêêD)ïå of ç
ü

êï3êë3/2êêè 3/2
# r = ─────────── = ─────────,ïε = 1,ïx = ─── = 3/2, parabola
ë2 + 2∙cos Θè1 + cos Θêêï1
Ç B
 7èGive the eccentricity, directrix, and name of the conic,
êêêër = 4/(12 + 15∙cos Θ).

êêêA)ïε = 1,ïx = 4/15,ïparabola
êêêB)ïε = 4/5,ïx = 4/15,ïellipse
êêêC)ïε = 5/4,ïx = 4/15,ïhyperbola
êêêD)ïå of ç
ü

êè 4êê1/3êêê 1/3
# r = ───────────── = ─────────────,ïε = 5/4,ïx = ─── = 4/15, hyperbola
ë12 + 15∙cos Θè1 + 5/4∙cos Θêêè5/4
Ç C
 8èGive the eccentricity, directrix, and name of the conic,
êêêêër = -2.5.

êêêA)ïNo graph
êêêB)ïε = 0,ïcircle
êêêC)ïline
êêêD)ïå of ç
ü


 This is the equation of a circle of radius 2.5.ïIf you use your ima-
gination, you could say that this conic has two foci that coincide and
a directrix at infinity.ïThus, the distance from the focus to a point
divided by the distance from the point to the directrix is zero.
Ç B
äïPlease give the eccentricity, the equation of the
êêdirectrix, and the name of each of the following conics.

âëGive the eccentricity, directrix, and name of the conic,
êêêër = 3/(2 - 2∙sin Θ).

êë3/2
#ër = ─────────,ëε = 1,ëy = -(3/2)/ε = -3/2,ëparabola
êï1 - sin Θ
éSïIn the previous problems of this section, the conics have
opened to the left or to the right.ïIn other words, the major axes of
the conics have been horizontal.ïIt is possible for conics to open up
or down, i.e. for their major axes to be vertical.ïIn the figure, the
#êêêëratio of │OP│ to │PT│ is ε = r/(r∙sin Θ + p).
êêêëSolving for r, r = ε∙p/(1 - ε∙sin Θ).ïThe
êêêëequation of the directrix is y = -p.
êêêê In a slightly different orientation with
êêêëthe conic opening down, the equation becomes
êêêër = ε∙p/(1 + ε∙sin Θ) with directrix y = p.
êêêëYou can see the similarities in the equations
@fig7502.bmp,10,100
êêêëof conics with major axes that are horizontal
êêêëand those with major axes that are vertical.
êêêëThe only difference is using the sin Θ instead
êêêëof the cos Θ, and the directrix is horizontal
êêêëinstead of vertical.
êêêêïOne last connection could be made, al-
êêêëthough it is not that helpful to do so.ïYou
êêêëcould use the identity, sin Θ = cos (π/2 - Θ),
êêêëto express all conics in just one polar equa-
êêêëtion form.
 9èGive the eccentricity, directrix, and name of the conic,
êêêër = 4/(2 - 2∙sin Θ).

êêêA)ïε = 1,ïy = -2,ïparabola
êêêB)ïε = 2,ïy = -2,ïhyperbola
êêêC)ïε = 1/2,ïy = -4,ïellipse
êêêD)ïå of ç
ü

êè 4êë 2
#èr = ─────────── = ─────────,ïε = 1,ïy = -2/1 = -2,ïparabola
ê2 - 2∙sin Θè1 - sin Θ
Ç A
 10ïGive the eccentricity, directrix, and name of the conic,
êêêër = 4/(6 + 3∙sin Θ).

êêêA)ïε = 1,ïy = 4/3,ïparabola
êêêB)ïε = 2,ïy = 3/4,ïhyperbola
êêêC)ïε = 1/2,ïy = 4/3,ïellipse
êêêD)ïå of ç
ü

êè 4êê 2/3êêêï2/3
#èr = ─────────── = ───────────────,ïε = 1/2,ïy = ─── = 4/3,ïellipse
ê6 + 3∙sin Θè1 + (1/2)∙sin Θêêè1/2
Ç C
 11ïGive the eccentricity, directrix, and name of the conic,
êêêër = 2/(3 - 4∙sin Θ).

êêêA)ïε = 2/3,ïy = -1/2,ïellipse
êêêB)ïε = 4/3,ïy = -1/2,ïhyperbola
êêêC)ïε = 1,ïy = -1/2,ïparabola
êêêD)ïå of ç
ü

êï2êê 2/3êêê -2/3
# r = ─────────── = ───────────────,ïε = 4/3,ïy = ─── = -1/2, hyperbola
ë3 - 4∙sin Θè1 - (4/3)∙sin Θêêè4/3
Ç B
 12ïGive the eccentricity, directrix, and name of the conic,
êêêêër = 5.

êêêA)ïNo graph
êêêB)ïε = 0,ïcircle
êêêC)ïline
êêêD)ïå of ç
ü


 This is the equation of a circle of radius 5.ïIf you use your imagina-
tion, you could say that this conic has two foci that coincide and a
directrix at infinity.ïThus, the distance from the focus to a point
divided by the distance from the point to the directrix is zero.
Ç B
 13ïGive the eccentricity, directrix, and name of the conic,
êêêër = 1/(4 - 3∙sin Θ).

êêêA)ïε = 3/4,ïy = -1/3,ïellipse
êêêB)ïε = 1,ïy = 3,ïparabola
êêêC)ïε = 4/3,ïy = -3,ïhyperbola
êêêD)ïå of ç
ü

êï1êê 1/4êêê -1/4
# r = ─────────── = ───────────────,ïε = 3/4,ïy = ─── = -1/3, ellipse
ë4 - 3∙sin Θè1 - (3/4)∙sin Θêêè3/4
Ç A
 14ïGive the eccentricity, directrix, and name of the conic,
êêêër = 3/(2 + 2∙sin Θ).

êêêA)ïε = 2/3,ïy = 1/2,ïellipse
êêêB)ïε = 1,ïy = 3/2,ïparabola
êêêC)ïε = 4/3,ïy = 3/4,ïhyperbola
êêêD)ïå of ç
ü

êï3êë3/2êêè 3/2
# r = ─────────── = ─────────,ïε = 1,ïy = ─── = 3/2, parabola
ë2 + 2∙sin Θè1 + sin Θêêï1
Ç B
 15ïGive the eccentricity, directrix, and name of the conic,
êêêër = 4/(12 + 15∙sin Θ).

êêêA)ïε = 1,ïy = 4/15,ïparabola
êêêB)ïε = 4/5,ïy = 4/15,ïellipse
êêêC)ïε = 5/4,ïy = 4/15,ïhyperbola
êêêD)ïå of ç
ü

êè 4êê1/3êêê 1/3
# r = ───────────── = ─────────────,ïε = 5/4,ïy = ─── = 4/15, hyperbola
ë12 + 15∙sin Θè1 + 5/4∙sin Θêêè5/4
Ç C
äïPlease draw the graphs of the following conics.
â

êêêëPlease see Details.
éSïTo draw the graph of r = 3/(2 - 2∙cos Θ), we would first find
the eccentricity, directrix, and the name of the conic.ïKnowing this
information will allow us to minimize the number of points we have to
plot in order to get a graph.ïBy writing the equation in the form,
r = (3/2)/(1 - cos Θ), we know ε = 1, x = -3/2, and that this is a pa-
rabola.ïAt this point we know the symmetry of the shape of a parabola,
êêêèand that it opens to the right as indicated
êêêèby the minus sign in front of cos Θ.ïIt
êêêèis generally sufficient to plot points for
êêêèvalues of Θ = 0, π/2, π, 3π/2, and 2π.
#êêêêïΘ │ï0ï│ π/2 │ïπï│ 3π/2 │ï2π
#êêêêï──┼─────┼─────┼─────┼──────┼─────
#êêêêïr │no pt│ 3/2 │ 3/4 │ 3/2ï│no pt
@fig7503.bmp,28,130
êêêèThese points are plotted on a polar coordinate
êêêèsystem to produce the graph of the parabola in
êêêèthe figure.
 16êïDraw a graph of r = 4/(2 - 2∙cos Θ).

@fig7504.bmp,28,225
@fig7505.bmp,28,234
@fig7508.bmp,28,450
üë First, find the eccentricity, directrix, and name.
êè4êë 2
#ïr = ─────────── = ─────────,ïε = 1,ïx = -2,ïparabola
ë 2 - 2∙cos Θè1 - cos Θ
#êêêêë Θ │ï0ï│ π/2 │ïπï│ 3π/2 │ï2π
#Then find ordered pairs forê──┼─────┼─────┼─────┼──────┼─────
#Θ = 0, π/2, π, 3π/2, and πê r │no pt│ï2ï│ï1ï│ï2è│no pt






@fig7505.bmp,3000,3100
Ç B
 17êïDraw a graph of r = 4/(6 + 3∙cos Θ).

@fig7506.bmp,28,225
@fig7507.bmp,28,234
@fig7508.bmp,28,450
üë First, find the eccentricity, directrix, and name.
êè4êë 2/3
#ïr = ─────────── = ─────────────,ïε = 1/2,ïx = 4/3,ïellipse
ë 6 + 3∙cos Θè1 + 1/2∙cos Θ
#êêêêë Θ │ï0ï│ π/2 │ïπï│ 3π/2 │ï2π
#Then find ordered pairs forê──┼─────┼─────┼─────┼──────┼─────
#Θ = 0, π/2, π, 3π/2, and πê r │ 4/9 │ 2/3 │ 4/3 │ 2/3ï│ 4/9






@fig7506.bmp,3000,3100
Ç A
 18êïDraw a graph of r = 2/(3 - 4∙cos Θ).

@fig7509.bmp,28,225
@fig7510.bmp,28,234
@fig7508.bmp,28,450
üêè First, ε = 4/3, x = -1/2, hyperbola.

#êêêêë Θ │ï0ï│ π/2 │ïπï│ 3π/2 │ï2π
#êêêêë ──┼─────┼─────┼─────┼──────┼─────
#êêêêë r │ -2ï│ 2/3 │ 2/7 │ 2/3ï│ -2

êêêêë Notice that we used symmetry to get
êêêêë the left half of the hyperbola.
@fig7509.bmp,1000,2000
Ç A