MTH120 College Algebra Chapter 8.5

 8.5 Conic Sections in Polar Coordinates

Conic sections can also be described and analyzed in polar coordinates, which use the distance (r) from the origin and the angle (θ) with respect to the positive x-axis to specify the location of a point. In polar coordinates, the equations for conic sections take on a different form. Here's how conic sections can be expressed in polar coordinates:

1. Polar Equation of a Circle:

The general equation of a circle in polar coordinates with its center at the origin is:

$�=�$

Where "a" is the radius of the circle.

2. Polar Equation of an Ellipse:

The general equation of an ellipse in polar coordinates with its center at the origin is:

$�=\frac{�(1-{�}^2)}{1-�\cos (�)}$

Where:

• "a" is the semi-major axis.
• "e" is the eccentricity (0 < e < 1).
• θ represents the angle with respect to the positive x-axis.

3. Polar Equation of a Hyperbola:

The general equation of a hyperbola in polar coordinates with its center at the origin is:

$�=\frac{�(1+{�}^2)}{1+�\cos (�)}$

Where:

• "a" is the distance from the origin to the closest point on the hyperbola.
• "e" is the eccentricity (e > 1).
• θ represents the angle with respect to the positive x-axis.

4. Polar Equation of a Parabola:

The general equation of a parabola in polar coordinates with its focus at the origin is:

$�=\frac{�}{1+\cos (�)}$

Where:

• "p" is the focal parameter, which determines the distance between the focus and the vertex.
• θ represents the angle with respect to the positive x-axis.

Analyzing conic sections in polar coordinates allows you to understand their shapes and characteristics using the radius "r" and the angle "θ" instead of Cartesian coordinates (x, y). This can be particularly useful when dealing with problems involving symmetrical or radial patterns.

Identifying a conic section in polar form involves recognizing the specific pattern of the polar equation and determining whether it represents a circle, ellipse, hyperbola, or parabola. Here are some examples to help you identify conic sections in polar form:

1. Polar Equation of a Circle: A polar equation of a circle has the form $�=�$ where "a" is a positive constant.

Example:

• $�=3$ represents a circle with a radius of 3.

2. Polar Equation of an Ellipse: A polar equation of an ellipse is generally of the form $�=\frac{�(1-{�}^2)}{1-�\cos (�)}$, where "a" is the semi-major axis and "e" is the eccentricity (0 < e < 1).

Example:

• $�=\frac{2(1-0.5^2)}{1-0.5\cos (�)}$ represents an ellipse with a semi-major axis of 2 and eccentricity of 0.5.

3. Polar Equation of a Hyperbola: A polar equation of a hyperbola is generally of the form $�=\frac{�(1+{�}^2)}{1+�\cos (�)}$, where "a" is the distance from the origin to the closest point on the hyperbola and "e" is the eccentricity (e > 1).

Example:

• $�=\frac{4(1+2^2)}{1+2\cos (�)}$ represents a hyperbola with a distance from the origin to the closest point of 4 and an eccentricity of 2.

4. Polar Equation of a Parabola: A polar equation of a parabola has the form $�=\frac{�}{1+\cos (�)}$, where "p" is the focal parameter.

Example:

• $�=\frac{3}{1+\cos (�)}$ represents a parabola with a focal parameter of 3.

To identify a conic section in polar form, you need to examine the structure of the equation and identify the key parameters like the semi-major axis, eccentricity, and focal parameter. These parameters help determine whether the polar equation represents a circle, ellipse, hyperbola, or parabola.

The polar equation for a conic section, whether it's a circle, ellipse, hyperbola, or parabola, describes the relationship between the radial distance (r) and the polar angle (θ) for points on the curve. Here are the polar equations for the different types of conics:

1. Polar Equation of a Circle:

For a circle with its center at the origin and radius "a," the polar equation is:

$�=�$

2. Polar Equation of an Ellipse:

For an ellipse centered at the origin with a semi-major axis "a," a semi-minor axis "b," and an eccentricity "e," the polar equation is:

$�=\frac{�(1-{�}^2)}{1-�\cos (�)}$

3. Polar Equation of a Hyperbola:

For a hyperbola centered at the origin with its transverse axis length "2a," eccentricity "e," and distance from the origin to the closest point on the hyperbola "c," the polar equation is:

$�=\frac{�(1+{�}^2)}{1+�\cos (�)}$

4. Polar Equation of a Parabola:

For a parabola with its focus at the origin and the focal parameter "p," the polar equation is:

$�=\frac{�}{1+\cos (�)}$

These equations express the relationship between the polar coordinates (r, θ) for points on the respective conic sections. By using these equations and manipulating the parameters, you can describe and analyze the shape and characteristics of conic sections in polar coordinates.

To identify the type of conic, the directrix, and the eccentricity from a polar equation, you need to examine the form of the equation and the specific parameters involved. Here are the steps to identify these characteristics:

1. Examine the Polar Equation:

• Carefully examine the polar equation you are given, looking for patterns that match the standard polar equations for conic sections.

2. Determine the Type of Conic:

• Based on the form of the equation, identify the type of conic section it represents:
  • If the equation is in the form $�=�$, it represents a circle.
  • If the equation is in the form $�=\frac{�(1-{�}^2)}{1-�\cos (�)}$, it represents an ellipse.
  • If the equation is in the form $�=\frac{�(1+{�}^2)}{1+�\cos (�)}$, it represents a hyperbola.
  • If the equation is in the form $�=\frac{�}{1+\cos (�)}$, it represents a parabola.

3. Determine the Eccentricity (for Ellipse and Hyperbola):

• If the equation represents an ellipse or hyperbola, calculate the eccentricity "e" using the relevant formula:
  • For an ellipse: $�=\sqrt{1-\frac{{�}^2}{{�}^2}}$, where "a" is the semi-major axis and "b" is the semi-minor axis.
  • For a hyperbola: $�=\sqrt{1+\frac{{�}^2}{{�}^2}}$, where "a" is the distance to the closest point on the hyperbola, and "c" is the distance from the origin to the center.

4. Determine the Directrix (for Parabola):

• If the equation represents a parabola, you can determine the directrix "D" based on the focal parameter "p" using the formula $�=\frac{�}{1+\cos (�)}$.
• The directrix is the line perpendicular to the polar axis (θ = 0) at a distance of "p" from the origin.

By following these steps, you can identify the type of conic section, calculate the eccentricity for ellipses and hyperbolas, and find the directrix for parabolas based on the given polar equation.

Conics can be defined in terms of a focus and a directrix, which are key elements in understanding their geometry. These definitions are particularly useful for parabolas, ellipses, and hyperbolas. Here's how each type of conic can be defined in terms of a focus and a directrix, along with examples:

1. Parabola:

Definition: A parabola is the set of all points that are equidistant from a fixed point called the focus (F) and a fixed line called the directrix (D). The distance from a point on the parabola to the focus is equal to the perpendicular distance to the directrix.

Example: Consider the parabola with focus (0, 2) and directrix y = -2. A point (x, y) on the parabola satisfies the condition that its distance to the focus (0, 2) is equal to the distance to the directrix y = -2. The equation that defines this parabola is:

$�=\frac{1}{2}(�-0{)}^2+2$

This is the standard form for a parabola with a vertex at the origin.

2. Ellipse:

Definition: An ellipse is the set of all points such that the sum of the distances from two fixed points, called the foci (F1 and F2), is constant. This constant sum is equal to the major axis length (2a).

Example: Consider an ellipse with foci at (0, 3) and (0, -3). The equation for this ellipse can be defined in terms of the distances from the foci:

$\sqrt{{�}^2+(�-3{)}^2}+\sqrt{{�}^2+(�+3{)}^2}=2�$

This equation defines an ellipse with a major axis along the y-axis.

3. Hyperbola:

Definition: A hyperbola is the set of all points such that the absolute difference of the distances from two fixed points, called the foci (F1 and F2), is constant. This constant difference is equal to the distance between the vertices (2a).

Example: Consider a hyperbola with foci at (-3, 0) and (3, 0). The equation for this hyperbola can be defined in terms of the differences between the distances to the foci:

$\mathrm{\mid }\sqrt{{�}^2+{�}^2}-3\mathrm{\mid }-\mathrm{\mid }\sqrt{{�}^2+{�}^2}+3\mathrm{\mid }=2�$

This equation defines a hyperbola with its transverse axis along the x-axis.

These definitions in terms of a focus and a directrix provide a geometrical insight into the properties of conic sections, making it easier to understand and work with them. The specific equations for conics can vary based on their orientation and location, but they all follow these general principles.

Algebra Chapter 8 Quiz

For the following exercises, write the equation in standard form and state the center, vertices, and foci. 1. x29+y24=1 2. 9y2+16x2−36y+32x−92=0 For the following exercises, sketch the graph, identifying the center, vertices, and foci. 3. (x−3)264+(y−2)236=1 4. 2x2+y2+8x−6y−7=0 5. Write the standard form equation of an ellipse with a center at (1,2), vertex at (7,2), and focus at (4,2). 6. A whispering gallery is to be constructed with a length of 150 feet. If the foci are to be located 20 feet away from the wall, how high should the ceiling be? For the following exercises, write the equation of the hyperbola in standard form, and give the center, vertices, foci, and asymptotes. 7. x249−y281=1 8. 16y2−9x2+128y+112=0 For the following exercises, graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes. 9. (x−3)225−(y+3)21=1 10. y2−x2+4y−4x−18=0 11. Write the standard form equation of a hyperbola with foci at (1,0) and (1,6), and a vertex at (1,2). For the following exercises, write the equation of the parabola in standard form, and give the vertex, focus, and equation of the directrix. 12. y2+10x=0 13. 3x2−11x−y+11=0 For the following exercises, graph the parabola, labeling the vertex, focus, and directrix. 14. (x−1)2=−4(y+3) 15. y2+8x−9y+40=0 16. Write the equation of a parabola with a focus at (2,3) and directrix y=−1. 17. A searchlight is shaped like a paraboloid of revolution. If the light source is located 1.5 feet from the base along the axis of symmetry, and the depth of the searchlight is 3 feet, what should the width of the opening be? For the following exercises, determine which conic section is represented by the given equation, and then determine the angle θ that will eliminate the xy term. 18. 3x2−2xy+5y2=4 19. x2+4xy+4y2+4x−8y=0 For the following exercises, rewrite in the x′y′ system without the x′y′ term, and graph the rotated graph. 20. 11x2+103–√xy+y2=4 21. 16x2+22xy+9y2−126x=0 For the following exercises, identify the conic with focus at the origin, and then give the directrix and eccentricity. 22. r=22−sinθ 23. r=24+6cosθ For the following exercises, graph the given conic section. If it is a parabola, label vertex, focus, and directrix. If it is an ellipse or a hyperbola, label vertices and foci. 24. r=125−8sinθ 25. r=22+4sinθ 26. Find a polar equation of the conic with focus at the origin, eccentricity of e=4, and directrix: x=2.

Let's work through these exercises one by one:

1. To write the equation ${�}^2\mathrm{/}9+{�}^2\mathrm{/}4=1$ in standard form:

• Divide both sides by 1 to isolate the terms on the left side.
• This is the equation of an ellipse centered at the origin with a major axis along the x-axis. Its standard form is $\frac{{�}^2}{{�}^2}+\frac{{�}^2}{{�}^2}=1$, where $�=3$ and $�=2$.
• The center is at (0, 0), and the vertices are at (±3, 0). The foci are at (±c, 0), where $�$ can be found using the relation $�=\sqrt{{�}^2-{�}^2}$.

2. First, rewrite the equation as $9{�}^2+16{�}^2-36�+32�-92=0$. Complete the square for both the x and y terms: $9({�}^2-4�)+16({�}^2+2�)=92$

To complete the square for the y-terms, add and subtract $(4\mathrm{/}2{)}^2=4$: $9({�}^2-4�+4)+16({�}^2+2�+1)=92+9(4)+16(1)$ $9(�-2{)}^2+16(�+1{)}^2=124$

Now, divide by 124 to obtain the standard form: $\frac{(�-2{)}^2}{(4\mathrm{/}3{)}^2}+\frac{(�+1{)}^2}{(\text{√}31\mathrm{/}3{)}^2}=1$

The center is at (-1, 2), and this is the equation of an ellipse. The major axis is along the x-axis, the minor axis is along the y-axis, the vertices can be found using $�=\text{√}(31\mathrm{/}3)$, and $�=4\mathrm{/}3$, and the foci can be found using $�=\text{√}({�}^2-{�}^2)$.

3. The equation $(�-3{)}^2\mathrm{/}64+(�-2{)}^2\mathrm{/}36=1$ is already in standard form for an ellipse. The center is at (3, 2), and the major axis is along the x-axis, with $�=8$ and $�=6$.

4. Rewrite the equation as $2({�}^2+4�)+({�}^2-6�)=7$. Completing the square for both the x and y terms: $2({�}^2+4�+4)+({�}^2-6�+9)=7+2(4)+9$ $2(�+2{)}^2+(�-3{)}^2=25$

Now, divide by 25 to obtain the standard form: $\frac{(�+2{)}^2}{(5\mathrm{/}2{)}^2}+\frac{(�-3{)}^2}{(5{)}^2}=1$

The center is at (-2, 3), and this is the equation of an ellipse. The major axis is along the y-axis, the minor axis is along the x-axis, the vertices can be found using $�=5$, and $�=5\mathrm{/}2$, and the foci can be found using $�=\text{√}({�}^2-{�}^2)$.

5. To write the standard form equation of an ellipse with a center at (1, 2), a vertex at (7, 2), and a focus at (4, 2), we can use the properties of an ellipse:

• The center is at (h, k) = (1, 2).
• The distance from the center to a vertex is "a," so $�=7-1=6$.
• The distance from the center to a focus is "c," so $�=4-1=3$.

Now, we can use the standard form equation of an ellipse with a center at (h, k), major axis along the x-axis, and major axis length 2a:

$\frac{(�-ℎ{)}^2}{{�}^2}+\frac{(�-�{)}^2}{{�}^2}=1$

Since the center is at (1, 2), and we already have the value of "a" as 6, we need to find "b." We can use the relationship ${�}^2={�}^2-{�}^2$ for ellipses:

$3^2=6^2-{�}^2$ $9=36-{�}^2$ ${�}^2=36-9$ ${�}^2=27$ $�=\sqrt{27}=3\sqrt{3}$

So, the standard form equation is:

$\frac{(�-1{)}^2}{36}+\frac{(�-2{)}^2}{27}=1$

6. The whispering gallery is an elliptical structure. To find the height of the ceiling (the semi-minor axis), we'll use the formula for the length of the major axis:

$2�=150$

Solving for "a":

$�=\frac{150}{2}=75$

Now, we know that "a" is 75, and you mentioned that the foci are 20 feet away from the wall. Since "c" is the distance from the center to a focus, we have $�=20$.

To find the value of "b" (the semi-minor axis), we can use the relationship ${�}^2={�}^2-{�}^2$ for ellipses:

$20^2=75^2-{�}^2$ $400=5625-{�}^2$ ${�}^2=5625-400$ ${�}^2=5225$ $�=\sqrt{5225}=5\sqrt{94}$

So, the height of the ceiling (the semi-minor axis) is $�=5\sqrt{94}$ feet.

7. To write the equation in standard form for a hyperbola:

• First, divide both sides by 1 to isolate the terms on the left side.
• Then, rewrite the equation as $\frac{{�}^2}{49}-\frac{{�}^2}{81}=1$.
• This is the equation of a hyperbola centered at the origin with a transverse axis along the x-axis. The standard form for a hyperbola is $\frac{{�}^2}{{�}^2}-\frac{{�}^2}{{�}^2}=1$, where $�$ and $�$ are the distances from the center to the vertices along the x and y-axes.

The center is at (0, 0), and the vertices are at (±7, 0). The foci can be found using $�=\sqrt{{�}^2+{�}^2}$, which is $�=\sqrt{49+81}=\sqrt{130}$. The asymptotes are the lines $�=\pm \frac{�}{�}�$, where $�\mathrm{/}�=9\mathrm{/}7$.

8. The equation $16{�}^2-9{�}^2+128�+112=0$ can be written in standard form for a hyperbola:

• Complete the square for both the x and y terms by adding and subtracting the necessary values: $16({�}^2+8�+16)-9({�}^2-112\mathrm{/}9)=0$
• This simplifies to $16(�+4{)}^2-9(�-16\mathrm{/}3{)}^2=144$.
• Divide by 144 to obtain the standard form: $\frac{(�+4{)}^2}{9}-\frac{(�-16\mathrm{/}3{)}^2}{16}=1$

The center is at (16/3, -4), and the vertices are at (±16/3, -4). The foci can be found using $�=\sqrt{{�}^2+{�}^2}$, where $�=4$ and $�=3$, so $�=5$. The asymptotes are the lines $�-(-4)=\pm \frac{�}{�}(�-16\mathrm{/}3)$, where $�\mathrm{/}�=3\mathrm{/}4$.

9. To write the equation $(�-3{)}^2\mathrm{/}25-(�+3{)}^2\mathrm{/}21=1$ in standard form for a hyperbola:

• Divide both sides by 1 to isolate the terms on the left side.
• The equation is already in standard form for a hyperbola centered at (3, -3) with a transverse axis along the x-axis. The standard form for this hyperbola is $\frac{(�-ℎ{)}^2}{{�}^2}-\frac{(�-�{)}^2}{{�}^2}=1$, where (h, k) is the center, "a" is the distance from the center to the vertices along the x-axis, and "b" is the distance from the center to the vertices along the y-axis.

The center is at (3, -3), and the vertices are at (±5, -3). The foci can be found using $�=\sqrt{{�}^2+{�}^2}$.

10. The equation ${�}^2-{�}^2+4�-4�-18=0$ can be written in standard form for a hyperbola:

• Complete the square for both the x and y terms.
• This can be rearranged as ${�}^2+4�-{�}^2-4�-18=0$.
• Completing the square for the y-terms and x-terms: ${�}^2+4�+4-{�}^2-4�-18+4=4$.
• Simplifying further, $(�+2{)}^2-(�+2{)}^2=4$.
• Divide by 4 to obtain the standard form: $\frac{(�+2{)}^2}{4}-\frac{(�+2{)}^2}{4}=1$.

The center is at (-2, -2), and this is the equation of a hyperbola. The standard form for a hyperbola with its center at (h, k) and a transverse axis along the x-axis is: $\frac{(�-ℎ{)}^2}{{�}^2}-\frac{(�-�{)}^2}{{�}^2}=1$, where (h, k) is the center, "a" is the distance from the center to the vertices along the x-axis, and "b" is the distance from the center to the vertices along the y-axis.

11. To write the standard form equation of a hyperbola with foci at (1, 0) and (1, 6), and a vertex at (1, 2):

• The center is at the midpoint of the foci, which is (1, (0 + 6)/2) = (1, 3).
• The distance from the center to a vertex is "a," which is the distance between (1, 3) and (1, 2), so $�=1$.
• The distance from the center to a focus is "c," which is 6/2 = 3.
• Use the standard form equation for a hyperbola centered at (h, k) with its transverse axis along the x-axis: $\frac{(�-ℎ{)}^2}{{�}^2}-\frac{(�-�{)}^2}{{�}^2}=1$.

So, you have $\frac{(�-1{)}^2}{1^2}-\frac{(�-3{)}^2}{{�}^2}=1$. To find "b," use the relationship ${�}^2={�}^2+{�}^2$, which gives $3^2=1^2+{�}^2$. Solving for "b," ${�}^2=9-1=8$, so $�=\sqrt{8}=2\text{√}2$.

The standard form equation is: $\frac{(�-1{)}^2}{1}-\frac{(�-3{)}^2}{(2\text{√}2{)}^2}=1$.

12. The equation ${�}^2+10�=0$ is already in standard form for a parabola with a vertex at the origin (0, 0), and the focus lies on the x-axis at (-5, 0). The equation of the directrix can be found using $�=-�$, where "a" is the distance from the vertex to the focus, so the directrix is $�=5$.

13. To write the equation $3{�}^2-11�-�+11=0$ in standard form for a parabola:

• Isolate the y-term by moving all other terms to the right side: $�=3{�}^2-11�+11$.
• The equation is in standard form for a parabola with a vertex at the origin (0, 0). The focus can be found by completing the square for the x-terms: $�=\frac{11}{2}=5.5$, so the focus is at (5.5, 0).
• The equation of the directrix is $�=-�$, where "a" is the distance from the vertex to the focus. So, the directrix is $�=-5.5$.

14. The equation $(�-1{)}^2=-4(�+3)$ is already in standard form for a parabola. This parabola has a vertex at (1, -3), and the focus lies on the y-axis at (1, -1). The equation of the directrix is $�=-5$.

15. To write the equation ${�}^2+8�-9�+40=0$ in standard form for a parabola:

• Rearrange the terms to isolate the y-terms: ${�}^2-9�=-8�-40$.
• Complete the square for the y-terms by adding and subtracting $(9\mathrm{/}2{)}^2=81\mathrm{/}4$: ${�}^2-9�+\frac{81}{4}=-8�-40+\frac{81}{4}$.
• Simplify: ${(�-\frac{9}{2})}^2=-8�-40+\frac{81}{4}$.

Now, this is in standard form for a parabola with its vertex at $(ℎ,�)$ and the focus on the x-axis. The vertex is $(\frac{9}{2},0)$, and the focus can be found by $�=-2$ (the coefficient of x), so the focus is at \left(\frac{9}{2} - 2, 0\right) = \left(\frac{5}{2}, 0\right). The equation of the directrix is \(x = -a, so the directrix is $�=2$.

16. To write the equation of a parabola with a focus at (2, 3) and a directrix $�=-1$:

• Since the focus is above the directrix, this is an upward-opening parabola.
• The vertex is the midpoint between the focus and the directrix, which is $(2,1)$.
• The distance from the vertex to the focus (or directrix) is the same, and it's the value of $�$.
• The value of $�$ is the vertical distance from the vertex to the focus (or directrix), which is $3-1=2$.
• The equation of the parabola in standard form is $(�-ℎ{)}^2=4�(�-�)$, where $(ℎ,�)$ is the vertex.

So, the equation is $(�-2{)}^2=8(�-1)$.

17. A paraboloid of revolution can be described by the equation $�=\frac{{�}^2+{�}^2}{4�}$, where "d" is the depth of the paraboloid. In this case, the light source is located 1.5 feet from the base, which is the value of "d." The depth is given as 3 feet.

To find the width of the opening, we need to determine the value of "r" in the equation z = \frac{x^2 + y^2}{4d}\ at the opening. The vertex of the paraboloid is at the origin (0, 0). At the opening, $�=3$ (the depth), so we have:

$3=\frac{{�}^2}{4(3)}$

Solving for "r":

$3=\frac{{�}^2}{12}$ ${�}^2=36$ $�=6\text{ feet}$

So, the width of the opening should be 6 feet.

18. The equation $3{�}^2-2��+5{�}^2=4$ is a conic section with both x and y terms. To eliminate the xy term, we need to rotate the axes.

Let $�$ be the angle of rotation. The trigonometric identity to use is $\tan (2�)=\frac{2�}{�-�}$, where A, B, and C are the coefficients of x^2, xy, and y^2 after the rotation. We want to find $�$ such that $�$ becomes 0.

In this case, $�=3$, $�=-2$, and $�=5$. Plug these values into the identity:

$\tan (2�)=\frac{2(-2)}{3-5}$ $\tan (2�)=\frac{-4}{-2}$ $\tan (2�)=2$

Now, find the angle $�$:

$�=\frac{1}{2}\arctan (2)$

The equation for the conic section after the rotation will be in terms of ${�}^{\mathrm{\prime }}$ and ${�}^{\mathrm{\prime }}$, where ${�}^{\mathrm{\prime }}=�\cos (�)-�\sin (�)$ and ${�}^{\mathrm{\prime }}=�\sin (�)+�\cos (�)$. This transformation eliminates the xy term.

19. The equation ${�}^2+4��+4{�}^2+4�-8�=0$ is a conic section with both x and y terms. To eliminate the xy term, we need to rotate the axes.

Follow the same steps as in exercise 18 to find the angle of rotation $�$. Once you find $�$, you can use the transformation equations ${�}^{\mathrm{\prime }}$ and ${�}^{\mathrm{\prime }}$ to rewrite the equation in terms of ${�}^{\mathrm{\prime }}$ and ${�}^{\mathrm{\prime }}$, effectively eliminating the xy term.

20. The equation $11{�}^2+10\sqrt{3}��+{�}^2=4$ represents an ellipse. To eliminate the xy term, we need to rotate the axes. Find the angle $�$ that eliminates the xy term using the same steps as in exercise 18.

Once you find $�$, you can use the transformation equations ${�}^{\mathrm{\prime }}$ and ${�}^{\mathrm{\prime }}$ to rewrite the equation in terms of ${�}^{\mathrm{\prime }}$ and ${�}^{\mathrm{\prime }}$, eliminating the xy term.

21. The equation $16{�}^2+22��+9{�}^2-126�=0$ represents an ellipse. To eliminate the xy term, we need to rotate the axes. Find the angle $�$ that eliminates the xy term using the same steps as in exercise 18.

Once you find $�$, you can use the transformation equations ${�}^{\mathrm{\prime }}$ and ${�}^{\mathrm{\prime }}$ to rewrite the equation in terms of ${�}^{\mathrm{\prime }}$ and ${�}^{\mathrm{\prime }}$, eliminating the xy term.

22. The polar equation $�=2-\sin (�)$ represents a conic section with its focus at the origin. To determine the conic type, calculate the eccentricity ($�$) using the formula $�=1-\frac{�}{�}$, where $�$ is the distance from the origin to the focus. In this case, $�=2$, and the origin is the focus.

$�=1-\frac{2-\sin (�)}{2}=1-(1-\frac{\sin (�)}{2})=\frac{\sin (�)}{2}$

Since $0\le �<1$ for all $�$, the conic is an ellipse.

23. The polar equation $�=4+6\cos (�)$ also represents a conic section with its focus at the origin. To determine the conic type, calculate the eccentricity ($�$) using the formula $�=1-\frac{�}{�}$, where $�$ is the distance from the origin to the focus. In this case, $�=4$, and the origin is the focus.

$�=1-\frac{4+6\cos (�)}{4}=1-(1+\frac{3\cos (�)}{2})=\frac{-3\cos (�)}{2}$

Since $0\le �<1$ for all $�$, the conic is an ellipse.

24. The polar equation $�=12-8\sin (�)$ represents a conic section with its focus at the origin. Calculate the eccentricity ($�$) using the formula $�=1-\frac{�}{�}$, where $�$ is the distance from the origin to the focus. In this case, $�=12$, and the origin is the focus.

$�=1-\frac{12-8\sin (�)}{12}=1-(1-\frac{2\sin (�)}{3})=\frac{2\sin (�)}{3}$

Since $0\le �<1$ for all $�$, the conic is an ellipse.

25. The polar equation $�=22+4\sin (�)$ also represents a conic section with its focus at the origin. Calculate the eccentricity ($�$) using the formula $�=1-\frac{�}{�}$, where $�$ is the distance from the origin to the focus. In this case, $�=22$, and the origin is the focus.

$�=1-\frac{22+4\sin (�)}{22}=1-(1+\frac{2\sin (�)}{11})=\frac{-2\sin (�)}{11}$

Since $0\le �<1$ for all $�$, the conic is an ellipse.

26. To find the polar equation of the conic with focus at the origin, eccentricity $�=4$, and directrix $�=2$, you can use the polar conic equation:

$�=\frac{��}{1+�\cos (�)}$

In this case, $�=4$ and the directrix is $�=2$, which means that the distance from the origin to the directrix (d) is 2. Now, plug these values into the equation:

$�=\frac{4\cdot 2}{1+4\cos (�)}$ $�=\frac{8}{1+4\cos (�)}$

So, the polar equation for this conic is $�=\frac{8}{1+4\cos (�)}$.