Precalculus by Richard Wright

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Nehemiah said, “Go and enjoy choice food and sweet drinks, and send some to those who have nothing prepared. This day is holy to our Lord. Do not grieve, for the joy of the Lord is your strength.” Nehemiah‬ ‭8‬:‭10‬ ‭NIV‬‬‬

7-09 Polar Graphs of Conics

Mr. Wright teaches the lesson.

Summary: In this section, you will:

SDA NAD Content Standards (2018): PC.5.2, PC.6.7

satellite in orbit
Figure 1: Aqua satellite in orbit above the earth. credit (NASA/JPL)

Satellites orbit the earth or other planets in ellipses with the planet at one focus. Because everything is space is moving in curved paths and contain round objects, it is logical to use a round coordinate system to describe space. It turns out that all conic sections can be described by one general polar equation.

Alternative Definition of a Conic Section

Locus of a point in the plane that moves so its distance from a fixed point (focus) is in a constant ratio to its distance from a fixed line (directrix).

The ratio is the eccentricity (e).

ellipse
If \(e < 1\), ellipse.
parabola
If \(e = 1\), parabola.
hyperbola
If \(e > 1\), hyperbola.
Polar Equation of Conic Sections

For all of these, a focus is at the pole.

Vertical Directrix
Directrix right of pole Directrix left of pole
$$ r = \frac{ep}{1+e \cos θ} $$ $$ r = \frac{ep}{1-e \cos θ} $$
right of pole left of pole


Horizontal Directrix
Directrix above pole Directrix below pole
$$ r = \frac{ep}{1+e \sin θ} $$ $$ r = \frac{ep}{1-e \sin θ} $$
above pole below pole

Example 1: Identify Type of Conic

Identify the type of conic of \(r = \frac{5}{2+4 \sin θ}\).

Solution

All the forms of a conic equation start with a 1 in the denominator. To get that, multiply the numerator and denominator by \(\frac{1}{2}\).

$$ r = \frac{5}{2+4 \sin θ} $$

$$ r = \frac{\frac{5}{2}}{1 + 2 \sin θ} $$

The coefficient of the trig function is e. In this case, e = 2 > 1 so it is a hyperbola. The denominator has a \(+ \sin θ\) which looks like \(r = \frac{ep}{1+e \sin θ}\). So the conic is a hyperbola with horizontal directrix above the pole.

r = (5)/(2+4 \sin θ)
Figure 2: \(r = \frac{5}{2+4 \sin θ}\)

Example 2: Identify Type of Conic

Identify the type of conic of \(r = \frac{5}{3-\cos θ}\) and then graph it.

Solution

All the forms of a conic equation start with a 1 in the denominator. To get that, multiply the numerator and denominator by \(\frac{1}{3}\).

$$ r = \frac{5}{3-\cos θ} $$

$$ r = \frac{\frac{5}{3}}{1 - \frac{1}{3} \cos θ} $$

The coefficient of the trig function is e. In this case, \(e = \frac{1}{3} < 1\) so it is a ellipse. The denominator has a \(- \cos θ\) which looks like \(r = \frac{ep}{1-e \cos θ}\). So the conic is a ellipse with vertical directrix left of the pole.

Graph the ellipse by making a table of values.

r 2.5 2.46 2.34 2.18 2 1.82 1.67 1.53 1.43 1.35 1.29 1.26 1.25
θ 0 \(\frac{π}{12}\) \(\frac{π}{6}\) \(\frac{π}{4}\) \(\frac{π}{3}\) \(\frac{5π}{12}\) \(\frac{π}{2}\) \(\frac{7π}{12}\) \(\frac{2π}{3}\) \(\frac{3π}{4}\) \(\frac{5π}{6}\) \(\frac{11π}{12}\) \(π\)
r 1.26 1.29 1.35 1.43 1.53 1.67 1.82 2 2.18 2.34 2.46 2.5
θ \(\frac{13π}{12}\) \(\frac{7π}{6}\) \(\frac{5π}{4}\) \(\frac{4π}{3}\) \(\frac{17π}{12}\) \(\frac{3π}{2}\) \(\frac{19π}{12}\) \(\frac{5π}{3}\) \(\frac{7π}{4}\) \(\frac{11π}{6}\) \(\frac{23π}{12}\) \(2π\)
r = (5)/(2+4 \sin θ)
Figure 3: \(r = \frac{5}{3- \cos θ}\)
Try It 1

Identify the type of conic of \(r = \frac{4}{1-\sin θ}\).

Answer

Parabola with horizontal directrix below the pole

Example 3: Find the Polar Equation of a Conic

Find the polar equation of the parabola whose focus is the pole and directrix is the line \(y = 3\).

Solution

Since the focus is the origin and the directrix is y = 3, the directrix is horizontal and above the pole. Thus, our equation is

$$ r = \frac{ep}{1 + e \sin θ} $$

Since it is a parabola, e = 1. p is the distance from the focus (0, 0) to the directrix y = 3, so p = 3. Fill those in and the equation becomes

$$ r = \frac{1(3)}{1 + 1 \sin θ} $$

$$ r = \frac{3}{1 + \sin θ} $$

r = 3/(1 + \sin θ)
Figure 4: \(r = \frac{3}{1 + \sin θ}\)

Example 4: Find the Polar Equation of a Conic

Find the polar equation of the ellipse whose focus is the pole and has vertices \(\left(2, 0\right)\) and \(\left(8, π\right)\).

Solution

Draw a quick sketch of the points and draw the primary axis of the ellipse. Remember the points are in polar format.

points
Figure 5: A plot of the vertices \(\left(2, 0\right)\) and \(\left(8, π\right)\).

The directrix will be to the right and the ellipse will bend away from it. So the directrix is vertical and to the right of the origin.

$$ r = \frac{ep}{1 + e \cos θ} $$

points
Figure 6: Add a and c to graph.

The center is halfway between the two vertices. a is the distance from the center to vertex, and c is the distance from the center to the focus. Remember from before that \(e = \frac{c}{a}\). In this case, \(a = 5\) and \(c = 3\), so \(e = \frac{3}{5}\). Fill this in and the equation becomes

$$ r = \frac{\frac{3}{5}p}{1 + \frac{3}{5} \cos θ} $$

Multiply numerator and denominator by 5 to remove fractions.

$$ r = \frac{3p}{5 + 3 \cos θ} $$

Now plug in a vertex point such as (2, 0) and solve for p.

$$ 2 = \frac{3p}{5 + 3 \cos 0} $$

$$ 2 = \frac{3p}{8} $$

$$ 16 = 3p $$

$$ \frac{16}{3} = p $$

Plug this into \(r = \frac{3p}{5 + 3 \cos θ}\) and simplify to get the final equation.

$$ r = \frac{3\left(\frac{16}{3}\right)}{5 + 3 \cos θ} $$

$$ r = \frac{16}{5 + 3 \cos θ} $$

r = \frac{16}{5 + 3 \cos θ}
Figure 7: \(r = \frac{16}{5 + 3 \cos θ}\)
Try It 2

Find the polar equation of the hyperbola whose focus is the pole and has vertices \(\left(1, \frac{3π}{2}\right)\) and \(\left(9, \frac{3π}{2}\right)\).

Answer

\(r = \frac{9}{4 - 5 \sin θ}\)

Lesson Summary

Alternative Definition of a Conic Section

Locus of a point in the plane that moves so its distance from a fixed point (focus) is in a constant ratio to its distance from a fixed line (directrix).

The ratio is the eccentricity (e).

ellipse
If \(e < 1\), ellipse.
parabola
If \(e = 1\), parabola.
hyperbola
If \(e > 1\), hyperbola.

Polar Equation of Conic Sections

For all of these, a focus is at the pole.

Vertical Directrix
Directrix right of pole Directrix left of pole
$$ r = \frac{ep}{1+e \cos θ} $$ $$ r = \frac{ep}{1-e \cos θ} $$
right of pole left of pole


Horizontal Directrix
Directrix above pole Directrix below pole
$$ r = \frac{ep}{1+e \sin θ} $$ $$ r = \frac{ep}{1-e \sin θ} $$
above pole below pole

Helpful videos about this lesson.

Practice Exercises

  1. Review the lessons on ellipses and hyperbolas. What does a, b, and c stand for? And how do you use them to find e?
  2. Identify the conic and sketch its graph.

  3. \(r = \frac{4}{2 + \sin θ}\)
  4. \(r = \frac{3}{1 - \cos θ}\)
  5. \(r = \frac{3}{2 - \sin θ}\)
  6. \(r = \frac{5}{1 + 4 \cos θ}\)
  7. \(r = \frac{5}{2 + 3 \cos θ}\)
  8. Use a graphing utility to graph the polar equation. Identify the graph.

  9. \(r = \frac{5}{2 + 2 \sin θ}\)
  10. \(r = \frac{5}{2 - \cos \left(θ - ^π/_4\right)}\)
  11. Write the polar equation of the conic with its focus at the pole and the given properties.

  12. parabola with directrix \(x = 4\)
  13. hyperbola with eccentricity \(e = 2\) and directrix \(y = -2\)
  14. ellipse with eccentricity \(e = \frac{2}{3}\) and directrix \(y = 6\)
  15. parabola with vertex \(\left(4, \frac{3π}{2}\right)\)
  16. ellipse with vertices \(\left(7, 0\right)\) and \(\left(3, π\right)\)
  17. hyperbola with vertices \(\left(3, π\right)\) and \(\left(−5, 0\right)\)
  18. Problem Solving

  19. The comet Hale-Bopp has an elliptical orbit with eccentricity of 0.995. It has a semi-major axis of about 250 astronomical units. (a) Write an equation for the orbit of Hale-Bopp with the sun at one focus. (b) How close does the comet come to the sun?
  20. Mixed Review

  21. (7-08) Graph \(r = 2 \sin θ\).
  22. (7-08) Identify the symmetry of \(r = 3 \cos θ\).
  23. (7-07) Convert \(r = 3 \cos θ\) to rectangular coordinates.
  24. (7-04) Find the standard equation of the hyperbola with foci (±10, 0) and vertices (±6, 0).
  25. (7-01) Find the angle between the lines \(y = \frac{2}{3}x + 1\) and \(y = x - 2\).

Answers

  1. a = distance between center and vertex; b = distance between center and covertex; c = distance between center and focus; \(e = \frac{c}{a}\)
  2. ellipse
    ans
  3. parabola
    ans
  4. ellipse
    ans
  5. hyperbola
    ans
  6. hyperbola
    ans
  7. parabola
    ans
  8. ellipse rotated by \(\frac{π}{4}\)
    ans
  9. \(r = \frac{4}{1 + \cos θ}\)
  10. \(r = \frac{4}{1 - 2\sin θ}\)
  11. \(r = \frac{12}{3 + 2\sin θ}\)
  12. \(r = \frac{8}{1 - \sin θ}\)
  13. \(r = \frac{21}{5 - 2\cos θ}\)
  14. \(r = \frac{15}{1 - 4\cos θ}\)
  15. \(r = \frac{2.494}{1 - 0.995 \cos θ}\); 1.25 AU
  16. ans
  17. Polar axis
  18. \(\left(x - \frac{3}{2}\right)^2 + y^2 = \frac{9}{4}\)
  19. \(\frac{x^2}{36} - \frac{y^2}{64} = 1\)
  20. 11.3°