Free math worksheets

- The directrix is vertical, so the polar equation uses a cosine term. - A directrix to the right of the pole produces a plus sign in the denominator. - Classify the conic by comparing the eccentricity with \(1\).

1. A vertical directrix to the right of the pole uses the form \(r=\frac{ed}{1+e\cos\theta}\). 2. Here \(ed=\frac{3}{4}(8)=6\). 3. Substitution gives \(r=\frac{6}{1+\frac{3}{4}\cos\theta}\). 4. Since \(0<e<1\), the conic is an ellipse.

\(r=\frac{6}{1+\frac{3}{4}\cos\theta}\); the conic is an ellipse.

- Compare the eccentricity with \(1\). - Recall how \(e<1\), \(e=1\), and \(e>1\) distinguish the three conic types.

1. A polar conic with \(0<e<1\) is an ellipse. 2. Since \(0.65<1\), the conic is an ellipse.

Ellipse

- A vertical directrix makes the perpendicular distance depend on the x-coordinate of the polar point. - Decide whether that x-coordinate is added to or subtracted from the positive directrix position. - Use the focus-directrix relation before solving for \(r\).

1. A vertical directrix uses the x-coordinate \(r\cos\theta\), so the angular term must involve \(\cos\theta\). 2. With the directrix at \(x=d>0\), the distance from a point on the conic to the directrix is \(d-r\cos\theta\). 3. The focus-directrix relation \(r=e(d-r\cos\theta)\) gives \(r(1+e\cos\theta)=ed\). 4. Therefore the denominator is \(1+e\cos\theta\).

A. \(1+e\cos\theta\)

- Compare the graph's distance from the pole in the positive and negative x-directions. - Test the two equations at \(\theta=0\) and \(\theta=\pi\). - The denominator sign determines which axial direction has the larger focal radius.

1. The graph extends farther to the left of the pole than to the right. 2. For \(r=\frac{3}{1+0.5\cos\theta}\), the radius at \(\theta=0\) is \(2\), while the radius at \(\theta=\pi\) is \(6\), so the curve extends farther left. 3. Reversing the sign would reverse those near and far directions. 4. Therefore equation A matches the graph.

A. \(r=\frac{3}{1+0.5\cos\theta}\)

- Normalize the denominator before identifying eccentricity and directrix distance. - Use the angle substitutions associated with reflections in polar coordinates. - Compare how sine changes under each substitution.

1. Rewrite the equation as \(r=\frac{3}{1+\frac{1}{2}\sin\theta}\). Thus \(e=\frac{1}{2}\), so the conic is an ellipse. 2. Since \(ed=3\), \(d=6\). The plus sine form corresponds to the directrix \(y=6\). 3. Reflection across the x-axis replaces \(\theta\) with \(-\theta\). Because \(\sin(-\theta)=-\sin\theta\), the reflected equation is \(r=\frac{6}{2-\sin\theta}\). 4. Reflection across the y-axis replaces \(\theta\) with \(\pi-\theta\). Since \(\sin(\pi-\theta)=\sin\theta\), the equation is unchanged.

a) Ellipse; directrix \(y=6\) b) \(r=\frac{6}{2-\sin\theta}\) c) Reflection across the y-axis leaves the graph unchanged.

- Multiply the polar equation by \(r\) before using the Cartesian substitutions. - Complete both squares to identify the circle's center and radius. - Set one coordinate equal to zero at a time to find the intercepts, and list the origin only once.

1. Multiply by \(r\): \(r^2=4r\cos\theta+4r\sin\theta\). 2. Substitute \(r^2=x^2+y^2\), \(r\cos\theta=x\), and \(r\sin\theta=y\): \(x^2+y^2=4x+4y\). 3. Complete the squares: \((x-2)^2+(y-2)^2=8\). 4. The conic is a circle centered at \((2,2)\) with radius \(2\sqrt{2}\). 5. On the x-axis, \(y=0\), so \(x^2=4x\), giving \(x=0\) or \(x=4\). 6. On the y-axis, \(x=0\), so \(y^2=4y\), giving \(y=0\) or \(y=4\). 7. The distinct axis intersections are \((0,0)\), \((4,0)\), and \((0,4)\).

Cartesian equation: \((x-2)^2+(y-2)^2=8\) Center: \((2, 2)\) Radius: \(2\sqrt{2}\) Axis intersections: \((0, 0)\), \((4, 0)\), and \((0, 4)\)

- Match the directrix orientation with the cosine form of the focus-directrix polar equation. - Use the eccentricity alone to classify the conic. - Evaluate the denominator at \(\cos0=1\) and \(\cos\pi=-1\).

1. For the directrix \(x=d\), the focus-directrix form is \(r=\frac{ed}{1+e\cos\theta}\). 2. Here \(ed=\frac{2}{3}(9)=6\), so \(r=\frac{6}{1+\frac{2}{3}\cos\theta}\). 3. Since \(0<e<1\), the conic is an ellipse. 4. At \(\theta=0\), \(r=\frac{6}{1+2/3}=\frac{18}{5}\). 5. At \(\theta=\pi\), \(r=\frac{6}{1-2/3}=18\).

a) \(r=\frac{6}{1+\frac{2}{3}\cos\theta}\) b) Ellipse c) \(r(0)=\frac{18}{5}\), \(r(\pi)=18\)

- Read each marked vertex's distance from the pole along the horizontal axis. - The near and far focal radii can be written as \(a-c\) and \(a+c\). - Use those two relationships to recover the ratio \(\frac{c}{a}\).

1. The near vertex is \(3\) units from the pole and the far vertex is \(9\) units from the pole. 2. For an ellipse measured from one focus along the major axis, \(r_{\text{near}}=a-c\) and \(r_{\text{far}}=a+c\). 3. Therefore \(a=\frac{3+9}{2}=6\) and \(c=\frac{9-3}{2}=3\). 4. The eccentricity is \(e=\frac{c}{a}=\frac{1}{2}\).

Near focal radius: \(3\) Far focal radius: \(9\) Eccentricity: \(e=\frac{1}{2}\)

- Decide whether \(\sin\theta\) represents the x-coordinate or y-coordinate of a polar point. - Compare the eccentricity with \(1\) before deciding the conic type. - Match the equation to the focus-directrix form for a horizontal directrix and track the sign of the directrix location carefully.

1. The eccentricity is \(e=0.5\), so the conic is an ellipse because \(e<1\). 2. A \(\sin\theta\) term comes from the y-coordinate \(r\sin\theta\), so the directrix is horizontal and the conic's axis is vertical. 3. Compare \(r=\frac{ed}{1-e\sin\theta}\). Since \(ed=4\) and \(e=0.5\), \(d=8\). 4. The denominator \(1-e\sin\theta\) corresponds to the directrix \(y=-d\), so the directrix is \(y=-8\). 5. Bastien confused \(\sin\theta\) with a vertical directrix and also interpreted the denominator sign in the opposite direction.

Conic: ellipse Axis: vertical Directrix: \(y=-8\) Errors: a \(\sin\theta\) term corresponds to a horizontal directrix because it uses the y-coordinate, and the minus sign here corresponds to the directrix below the pole, not above it.

- First rewrite the denominator so its constant term is \(1\). - Use the coefficient of the trigonometric term to decide the conic type. - Evaluate the equation in the two opposite directions along the polar axis.

1. Rewrite the equation as \(r=\frac{4}{1+\frac{2}{3}\cos\theta}\). Thus the eccentricity is \(e=\frac{2}{3}\), so the conic is an ellipse. 2. Comparing the numerator with the focus-directrix form gives \(ed=4\). Therefore \(d=6\), and the directrix is \(x=6\). 3. Along the positive polar axis, \(\theta=0\), so \(r=\frac{12}{5}\). 4. Along the negative polar axis, \(\theta=\pi\), so \(r=12\). 5. Therefore the nearest and farthest distances from the pole are \(\frac{12}{5}\) and \(12\), respectively.

a) The conic is an ellipse with eccentricity \(e=\frac{2}{3}\). b) The directrix is \(x=6\). c) The nearest distance is \(\frac{12}{5}\), and the farthest distance is \(12\).

- Normalize the denominator so its constant term is \(1\). - Use the sign of the sine term to determine which horizontal directrix is involved. - Directions of unbounded growth occur where the denominator approaches zero.

1. Rewrite the equation as \(r=\frac{5}{1-\frac{3}{2}\sin\theta}\). Thus \(e=\frac{3}{2}>1\), so the conic is a hyperbola. 2. The numerator equals \(ed=5\), giving \(d=\frac{10}{3}\). 3. The form \(1-e\sin\theta\) corresponds to the directrix \(y=-d\), so the directrix is \(y=-\frac{10}{3}\). 4. The radius becomes unbounded when \(2-3\sin\theta=0\), or \(\sin\theta=\frac{2}{3}\). 5. The two direction angles are \(\theta=\arcsin(\frac{2}{3})\) and \(\theta=\pi-\arcsin(\frac{2}{3})\).

a) Hyperbola with eccentricity \(e=\frac{3}{2}\) b) Directrix: \(y=-\frac{10}{3}\) c) \(\theta=\arcsin(\frac{2}{3})\) and \(\theta=\pi-\arcsin(\frac{2}{3})\)

- A line through the pole corresponds to two polar angles differing by \(\pi\). - Evaluate the radius separately in those opposite directions. - Convert each polar point to Cartesian coordinates and consider whether the pole lies between them.

1. One endpoint lies on the ray \(\theta=\frac{\pi}{3}\), where \(r=\frac{8}{2+\frac{1}{2}}=\frac{16}{5}\). 2. Its Cartesian coordinates are \((\frac{16}{5}\cos\frac{\pi}{3}, \frac{16}{5}\sin\frac{\pi}{3})=(\frac{8}{5}, \frac{8\sqrt{3}}{5})\). 3. The opposite endpoint lies on \(\theta=\frac{4\pi}{3}\), where \(r=\frac{8}{2-\frac{1}{2}}=\frac{16}{3}\). 4. Its coordinates are \((-\frac{8}{3}, -\frac{8\sqrt{3}}{3})\). 5. The pole lies between the endpoints, so the chord length is the sum of their radial distances: \(\frac{16}{5}+\frac{16}{3}=\frac{128}{15}\).

Endpoints: \((\frac{8}{5}, \frac{8\sqrt{3}}{5})\) and \((-\frac{8}{3}, -\frac{8\sqrt{3}}{3})\) Chord length: \(\frac{128}{15}\)

- At an intersection, the two formulas produce the same radial distance in a common direction. - Solve first for the trigonometric value rather than for the angle itself. - Convert the resulting polar coordinates to Cartesian coordinates at the end.

1. At a common point represented by the same angle, set the radii equal: \(\frac{6}{1+\frac{1}{2}\cos\theta}=\frac{4}{1-\frac{1}{2}\cos\theta}\). 2. Cross-multiplying gives \(6-3\cos\theta=4+2\cos\theta\), so \(\cos\theta=\frac{2}{5}\). 3. There are two angles in \([0,2\pi)\) with this cosine, and their sine values are \(\pm\frac{\sqrt{21}}{5}\). 4. Substitution into either equation gives \(r=5\). 5. Therefore \(x=r\cos\theta=2\) and \(y=r\sin\theta=\pm\sqrt{21}\).

\((2, \sqrt{21})\) and \((2, -\sqrt{21})\)

- Convert the full focal chord length into the numerator used in polar conic form. - Evaluate the equation in the direction of the nearest vertex. - Use the opposite polar-axis direction for the farthest point.

1. In \(r=\frac{\ell}{1+e\cos\theta}\), \(\ell\) is the semilatus rectum. A latus rectum of length \(10\) gives \(\ell=5\). 2. At the nearest vertex, \(\theta=0\), so \(4=\frac{5}{1+e}\). 3. Therefore \(1+e=\frac{5}{4}\), giving \(e=\frac{1}{4}\). Since \(e<1\), the conic is an ellipse. 4. Because \(\ell=ed\), the directrix distance is \(d=\frac{5}{1/4}=20\), so the directrix is \(x=20\). 5. The polar equation is \(r=\frac{5}{1+\frac{1}{4}\cos\theta}\). 6. The farthest point occurs at \(\theta=\pi\), where \(r=\frac{5}{1-1/4}=\frac{20}{3}\).

Ellipse with eccentricity \(e=\frac{1}{4}\) Directrix: \(x=20\) Polar equation: \(r=\frac{5}{1+\frac{1}{4}\cos\theta}\) Farthest distance from the pole: \(\frac{20}{3}\)

- Read eccentricity and directrix distance from normalized polar form. - A negative radius reverses the plotted direction by \(\pi\). - Once the axial vertices are known, use symmetry to locate the center and second focus.

1. The equation has eccentricity \(e=2>1\), so it represents a hyperbola. 2. Since \(ed=6\), \(d=3\). The minus cosine form corresponds to the directrix \(x=-3\). 3. At \(\theta=0\), \(r=-6\). A negative radius places the point \(6\) units in the opposite direction, so the Cartesian point is \((-6, 0)\). 4. At \(\theta=\pi\), \(r=2\), which also gives the Cartesian point \((-2, 0)\). 5. These are the two vertices, so their midpoint is the center \((-4, 0)\), and \(a=2\). 6. The pole \((0, 0)\) is one focus. Reflection across the center gives the second focus \((-8, 0)\).

a) Hyperbola; directrix \(x=-3\) b) \(\theta=0\) gives \((-6, 0)\); \(\theta=\pi\) gives \((-2, 0)\) c) Center \((-4, 0)\); vertices \((-6, 0)\), \((-2, 0)\); second focus \((-8, 0)\)

- Use the vertices to find the center and semi-major axis. - The given focus determines the focal distance and the second focus by symmetry. - Pair the pole focus with the directrix on the same side of the center.

1. The center is the midpoint of the vertices, \((3, 0)\), and the semi-major axis is \(a=5\). 2. Since one focus is \((0, 0)\), the focal distance is \(c=3\), so the other focus is \((6, 0)\). 3. The eccentricity is \(e=\frac{c}{a}=\frac{3}{5}\). 4. The directrix associated with the left focus is \(x=h-\frac{a}{e}=3-\frac{25}{3}=-\frac{16}{3}\). 5. Its distance from the pole is \(d=\frac{16}{3}\), so \(ed=\frac{16}{5}\). 6. For a left-hand directrix, \(r=\frac{ed}{1-e\cos\theta}\), giving \(r=\frac{16}{5-3\cos\theta}\).

Other focus: \((6, 0)\) Associated directrix: \(x=-\frac{16}{3}\) Polar equation: \(r=\frac{16}{5-3\cos\theta}\)

- Normalize the denominator while preserving the shifted angle. - Interpret the shifted cosine as projection onto a rotated axis. - Extreme radii occur along that axis and its opposite direction.

1. Rewrite the equation as \(r=\frac{9/2}{1+\frac{1}{2}\cos(\theta-\frac{\pi}{6})}\). Thus \(e=\frac{1}{2}\), so the conic is an ellipse. 2. Since \(ed=\frac{9}{2}\), the directrix distance is \(d=9\). 3. The directrix is perpendicular to the direction \(\alpha=\frac{\pi}{6}\), so its equation is \(x\cos\alpha+y\sin\alpha=9\). 4. Therefore the directrix is \(\frac{\sqrt{3}}{2}x+\frac{1}{2}y=9\), or \(\sqrt{3}x+y=18\). 5. The nearest distance occurs when \(\theta=\frac{\pi}{6}\): \(r=\frac{9}{3}=3\). 6. The farthest distance occurs in the opposite direction, \(\theta=\frac{7\pi}{6}\): \(r=\frac{9}{1}=9\).

Ellipse with eccentricity \(\frac{1}{2}\) Directrix: \(\sqrt{3}x+y=18\) Nearest distance: \(3\) at \(\theta=\frac{\pi}{6}\) Farthest distance: \(9\) at \(\theta=\frac{7\pi}{6}\)

- Start by evaluating the two directions along the polar axis rather than converting the full equation. - Treat those axial focal radii as the distances from one focus to the two vertices. - After finding the center and \(a\), use eccentricity to recover \(c\) and then the remaining semi-axis.

1. Rewrite the equation as \(r=\frac{4}{1+\frac{1}{3}\cos\theta}\), so \(e=\frac{1}{3}\). 2. Along \(\theta=0\), \(r=3\), giving the vertex \((3,0)\). Along \(\theta=\pi\), \(r=6\), giving the vertex \((-6,0)\). 3. The center is the midpoint of the vertices, \(\left(-\frac{3}{2},0\right)\), and \(a=\frac{9}{2}\). 4. The pole is one focus. Since \(c=ea=\frac{3}{2}\), the other focus is \((-3,0)\). 5. Then \(b^2=a^2-c^2=\frac{81}{4}-\frac{9}{4}=18\). 6. The standard equation is \(\frac{(x+\frac{3}{2})^2}{81/4}+\frac{y^2}{18}=1\).

a) \(r(0)=3\), \(r(\pi)=6\); vertices: \((3,0)\), \((-6,0)\); center: \(\left(-\frac{3}{2},0\right)\); \(a=\frac{9}{2}\) b) \(e=\frac{1}{3}\); foci: \((0,0)\), \((-3,0)\) c) \(b=3\sqrt{2}\); \(\frac{(x+\frac{3}{2})^2}{81/4}+\frac{y^2}{18}=1\)

- Start with the angle whose cosine is zero to isolate the constant term. - Use a second measurement to determine the cosine coefficient. - Test the remaining measurement before interpreting the two coefficients geometrically.

1. At \(\theta=\frac{\pi}{2}\), \(\cos\theta=0\), so \(A=\frac{1}{4}\). 2. At \(\theta=0\), \(\frac{1}{3}=A+B\), giving \(B=\frac{1}{12}\). 3. The model predicts \(\frac{1}{r(\pi)}=A-B=\frac{1}{4}-\frac{1}{12}=\frac{1}{6}\), which agrees with \(r(\pi)=6\). The measurements are consistent. 4. Comparing \(\frac{1}{r}=\frac{1}{\ell}+\frac{e}{\ell}\cos\theta\) gives \(\ell=4\) and \(e=\frac{B}{A}=\frac{1}{3}\). 5. Since \(e<1\), the conic is an ellipse. Also, \(\ell=ed\), so \(d=\frac{4}{1/3}=12\); the directrix is \(x=12\). 6. The polar equation is \(r=\frac{4}{1+\frac{1}{3}\cos\theta}\).

The measurements are consistent. Eccentricity: \(e=\frac{1}{3}\) Classification: ellipse Directrix: \(x=12\) Polar equation: \(r=\frac{4}{1+\frac{1}{3}\cos\theta}\)

- Replace \(r\sin\theta\) with a Cartesian coordinate. - Isolate \(r\) before using \(r^2=x^2+y^2\). - Compare the simplified equation with a standard parabola form.

1. Multiply by the denominator to get \(r+r\sin\theta=8\). 2. Since \(r\sin\theta=y\), this becomes \(r=8-y\). 3. Square both sides using \(r^2=x^2+y^2\): \(x^2+y^2=(8-y)^2\). 4. Simplifying gives \(x^2=-16(y-4)\). 5. Comparing with \((x-h)^2=4p(y-k)\) gives \((h, k)=(0, 4)\) and \(p=-4\). 6. Therefore, the focus is \((0, 0)\), the directrix is \(y=8\), and the parabola opens downward.

Cartesian equation: \(x^2=-16(y-4)\) Vertex: \((0, 4)\) Focus: \((0, 0)\) Directrix: \(y=8\) Opening direction: downward

- Replace \(\theta\) by \(\theta+\pi\) to write the opposite-ray distance. - Add the reciprocals before substituting any specific distance. - Use the given radius to determine the cosine of the direction angle.

1. The opposite-ray distances are \(r_1=\frac{5}{1+\frac{1}{2}\cos\theta}\) and \(r_2=\frac{5}{1-\frac{1}{2}\cos\theta}\). 2. Therefore, \(\frac{1}{r_1}+\frac{1}{r_2}=\frac{1+\frac{1}{2}\cos\theta}{5}+\frac{1-\frac{1}{2}\cos\theta}{5}=\frac{2}{5}\). 3. If \(r_1=4\), then \(\frac{5}{1+\frac{1}{2}\cos\theta}=4\), so \(\cos\theta=\frac{1}{2}\). 4. In \([0,2\pi)\), this gives \(\theta=\frac{\pi}{3}\) or \(\theta=\frac{5\pi}{3}\). 5. Using the reciprocal relation, \(\frac{1}{4}+\frac{1}{r_2}=\frac{2}{5}\), so \(r_2=\frac{20}{3}\).

a) \(\frac{1}{r_1}+\frac{1}{r_2}=\frac{2}{5}\) b) \(\theta=\frac{\pi}{3}\) or \(\theta=\frac{5\pi}{3}\), and \(r_2=\frac{20}{3}\)

- Combine the sine and cosine terms into one shifted trigonometric function. - Read the eccentricity from the normalized denominator. - Use the direction of the shifted cosine to orient the directrix.

1. Use \(\sin\theta-\sqrt{3}\cos\theta=2\cos\left(\theta-\frac{5\pi}{6}\right)\). 2. The equation becomes \(r=\frac{3}{1+\cos\left(\theta-\frac{5\pi}{6}\right)}\). 3. This has eccentricity \(e=1\), so the conic is a parabola with focus at the pole. 4. The focal-axis unit vector is \(\left(\cos\frac{5\pi}{6}, \sin\frac{5\pi}{6}\right)=\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\). 5. Since the semilatus rectum is \(3\) and \(e=1\), the directrix is \(3\) units from the pole in that direction: \(-\frac{\sqrt{3}}{2}x+\frac{1}{2}y=3\), or \(-\sqrt{3}x+y=6\). 6. The vertex is halfway from the focus to the directrix along the focal axis, so it is \(\left(-\frac{3\sqrt{3}}{4}, \frac{3}{4}\right)\).

Standard polar form: \(r=\frac{3}{1+\cos\left(\theta-\frac{5\pi}{6}\right)}\) Conic: parabola Directrix: \(-\sqrt{3}x+y=6\) Vertex: \(\left(-\frac{3\sqrt{3}}{4}, \frac{3}{4}\right)\)

- Relate the old angle to the angle measured from the rotated polar axis. - Expand the cosine of an angle sum. - Rotate the directrix equation using the same coordinate relationship.

1. The old and new angles satisfy \(\theta=\phi+\frac{\pi}{3}\). 2. Substitute and expand: \(\cos\left(\phi+\frac{\pi}{3}\right)=\frac{1}{2}\cos\phi-\frac{\sqrt{3}}{2}\sin\phi\). 3. The new polar equation is \(r=\frac{8}{1+\frac{1}{6}\cos\phi-\frac{\sqrt{3}}{6}\sin\phi}\). 4. The original equation has eccentricity \(\frac{1}{3}\) and semilatus rectum \(8\), so its associated directrix is \(x=24\). 5. The coordinate relation is \(x=x'\cos60^\circ-y'\sin60^\circ=\frac{1}{2}x'-\frac{\sqrt{3}}{2}y'\). 6. Therefore the directrix in the new coordinates is \(\frac{1}{2}x'-\frac{\sqrt{3}}{2}y'=24\), or \(x'-\sqrt{3}y'=48\).

New polar equation: \(r=\frac{8}{1+\frac{1}{6}\cos\phi-\frac{\sqrt{3}}{6}\sin\phi}\) Directrix in the new coordinates: \(x'-\sqrt{3}y'=48\)

- Substitute polar coordinates into the Cartesian circle equation without solving for \(r\) first. - Combine the discriminant condition with the sign of the sum of the radial roots. - At an endpoint of the angle interval, use the repeated-root formula to find \(r\) and then convert to Cartesian coordinates.

1. Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\): \(r^2-10r\cos\theta+16=0\). 2. Real intersections require the discriminant to be nonnegative: \(100\cos^2\theta-64\ge0\), so \(|\cos\theta|\ge\frac{4}{5}\). 3. For the roots to be nonnegative, their sum \(10\cos\theta\) must be positive. Therefore, \(\cos\theta\ge\frac{4}{5}\). 4. The intersecting ray directions are \(0\le\theta\le\arccos\left(\frac{4}{5}\right)\) or \(2\pi-\arccos\left(\frac{4}{5}\right)\le\theta<2\pi\). 5. At either boundary direction, the discriminant is zero and the repeated radius is \(r=5\cos\theta=4\). 6. The repeated intersection points are \(\left(\frac{16}{5},\frac{12}{5}\right)\) and \(\left(\frac{16}{5},-\frac{12}{5}\right)\).

a) \(r^2-10r\cos\theta+16=0\) b) \(0\le\theta\le\arccos\left(\frac{4}{5}\right)\) or \(2\pi-\arccos\left(\frac{4}{5}\right)\le\theta<2\pi\) c) \(\left(\frac{16}{5}, \frac{12}{5}\right)\) and \(\left(\frac{16}{5}, -\frac{12}{5}\right)\)

- Put the polar equation into the standard numerator-over-\(1\pm e\cos\theta\) form first. - Use the required semilatus-rectum relation before computing either Cartesian semi-axis. - Once both semi-axes are known, use the ordinary ellipse area formula.

1. Normalize the equation: \(r=\frac{6}{1+\frac{1}{3}\cos\theta}\). 2. Thus \(\ell=6\) and \(e=\frac{1}{3}\). 3. Using \(\ell=a(1-e^2)\), \(a=\frac{6}{1-1/9}=\frac{27}{4}\). 4. The semi-minor axis is \(b=a\sqrt{1-e^2}=\frac{27}{4}\sqrt{\frac{8}{9}}=\frac{9\sqrt{2}}{2}\). 5. The area is \(\pi ab=\pi\left(\frac{27}{4}\right)\left(\frac{9\sqrt{2}}{2}\right)=\frac{243\sqrt{2}\pi}{8}\).

a) \(\ell=6\), \(e=\frac{1}{3}\) b) \(a=\frac{27}{4}\), \(b=\frac{9\sqrt{2}}{2}\) c) \(\frac{243\sqrt{2}\pi}{8}\)

- Set the two radial equations equal. - Determine when the resulting cosine value is possible. - Separate the endpoint cosine value from values strictly between \(-1\) and \(1\).

1. At an intersection, \(R=\frac{8}{1+\cos\theta}\), so \(\cos\theta=\frac{8}{R}-1\). 2. A real angle exists when \(-1\le\frac{8}{R}-1\le1\). 3. Since \(R>0\), this condition reduces to \(R\ge4\). 4. If \(0<R<4\), there are no intersections. 5. If \(R=4\), then \(\cos\theta=1\), giving the single point \((r, \theta)=(4, 0)\). The circle is tangent to the parabola at its vertex. 6. If \(R>4\), then \(-1<\frac{8}{R}-1<1\), so there are two distinct intersection angles and two intersection points.

\(0<R<4\): no intersections \(R=4\): one intersection, with tangency at \((r, \theta)=(4, 0)\) \(R>4\): two intersections

- Substitute the polar coordinate definitions before isolating the radial variable. - A real radial distance requires a sign condition on the resulting denominator. - Compare the limiting angular directions with the slopes obtained from the Cartesian equation.

1. Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\): \(r^2\left(\frac{\cos^2\theta}{9}-\frac{\sin^2\theta}{16}\right)=1\). 2. Solving for \(r^2\) gives \(r^2=\frac{144}{16\cos^2\theta-9\sin^2\theta}\). 3. Real finite points require \(16\cos^2\theta-9\sin^2\theta>0\), which is equivalent to \(|\tan\theta|<\frac{4}{3}\), modulo \(\pi\). 4. At the boundary directions, \(\tan\theta=\pm\frac{4}{3}\), the denominator approaches zero and \(|r|\) grows without bound. 5. These boundary rays have equations \(y=\pm\frac{4}{3}x\), exactly the asymptotes of the hyperbola.

Polar form: \(r^2=\frac{144}{16\cos^2\theta-9\sin^2\theta}\) Real finite points occur for \(|\tan\theta|<\frac{4}{3}\), modulo \(\pi\). The boundary directions \(\tan\theta=\pm\frac{4}{3}\) are the asymptote directions.

- Convert each equation to Cartesian form to see the circle geometry. - Check the pole separately before equating nonzero radial expressions. - A useful direction ratio appears when the two radial expressions are equal.

1. Multiplying each equation by \(r\) gives \(x^2+y^2=6x\) and \(x^2+y^2=8y\). 2. The first circle is \((x-3)^2+y^2=9\), with center \((3, 0)\) and radius \(3\). The second is \(x^2+(y-4)^2=16\), with center \((0, 4)\) and radius \(4\). 3. The pole \((0, 0)\) lies on both circles, although it occurs at different angle values in the two polar equations. 4. For a nonzero-radius intersection, equating the polar expressions gives \(6\cos\theta=8\sin\theta\), so \(\tan\theta=\frac34\). 5. Using \(\cos\theta=\frac45\) and \(\sin\theta=\frac35\) gives \(r=\frac{24}{5}\), hence \((x, y)=\left(\frac{96}{25}, \frac{72}{25}\right)\).

First circle: center \((3, 0)\), radius \(3\) Second circle: center \((0, 4)\), radius \(4\) Intersection points: \((0, 0)\) and \(\left(\frac{96}{25}, \frac{72}{25}\right)\)

- Write both distances directly in polar coordinates. - Treat the pole separately before canceling the radial factor. - Interpret the remaining angular condition as rays in the coordinate plane.

1. A polar point has distance \(|r|\) from the pole and distance \(|r\cos\theta|\) from the line \(x=0\). 2. The focus-directrix condition is \(|r|=2|r\cos\theta|\). 3. The pole satisfies the condition. For a nonzero point, cancel \(|r|\) to obtain \(|\cos\theta|=\frac12\). 4. The allowed directions are \(\theta=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\). 5. These four rays form the two complete lines \(y=\sqrt3x\) and \(y=-\sqrt3x\). The locus is therefore a degenerate pair of intersecting lines.

The locus is \(y=\sqrt3x\) together with \(y=-\sqrt3x\), including the pole. It is not a nondegenerate hyperbola because placing the directrix through the focus collapses the radial condition to fixed directions rather than a curved branch.

- Use the eccentricity and semilatus rectum to locate the ellipse center relative to the focus at the pole. - Symmetric angles produce a vertical chord with a common x-coordinate. - Set that x-coordinate equal to the center's x-coordinate, then use the resulting reference triangle for the endpoint coordinates.

1. Here the semilatus rectum is \(10\) and \(e=\frac{3}{5}\), so \(a=\frac{10}{1-e^2}=\frac{125}{8}\). 2. The center is \((-ae,0)=\left(-\frac{75}{8},0\right)\). 3. The symmetric points have the same radius and x-coordinate \(x=r\cos\alpha\), so their chord is vertical. 4. Require the common x-coordinate to equal the center's x-coordinate: \(\frac{10\cos\alpha}{1+\frac{3}{5}\cos\alpha}=-\frac{75}{8}\). 5. Solving gives \(\cos\alpha=-\frac{3}{5}\), so \(\alpha=\arccos\left(-\frac{3}{5}\right)\). 6. At this angle, \(r=\frac{125}{8}\), and \(\sin\alpha=\frac{4}{5}\). 7. The endpoints are \(\left(-\frac{75}{8},\frac{25}{2}\right)\) and \(\left(-\frac{75}{8},-\frac{25}{2}\right)\). The chord passes through the center and is perpendicular to the horizontal major axis, so it is the minor axis.

\(\alpha=\arccos\left(-\frac{3}{5}\right)\) Endpoints: \(\left(-\frac{75}{8}, \frac{25}{2}\right)\) and \(\left(-\frac{75}{8}, -\frac{25}{2}\right)\) The chord is the minor axis.

- In part a), keep both curves in polar form and use their common angle and radius. - Check every algebraic value of \(\cos\theta\) against the allowed interval before recovering the angles. - Only after reporting the polar intersections should you use \(r^2=x^2+y^2\) and \(r\cos\theta=x\) for Cartesian verification.

1. At a common non-pole point, equate the radial expressions: \(\frac{8}{1+\frac{2}{3}\cos\theta}=12\cos\theta\). 2. Let \(c=\cos\theta\). Then \(8=12c+8c^2\), so \(2c^2+3c-2=0\). 3. The roots are \(c=\frac{1}{2}\) and \(c=-2\). Only \(c=\frac{1}{2}\) is possible for a cosine. 4. Therefore \(\theta=\pm\frac{\pi}{3}\), and \(r=12\left(\frac{1}{2}\right)=6\). The polar intersections are \(\left(6,\frac{\pi}{3}\right)\) and \(\left(6,-\frac{\pi}{3}\right)\). 5. Converting gives \((r\cos\theta,r\sin\theta)=(3,\pm3\sqrt{3})\). 6. For the ellipse, \(r+\frac{2}{3}r\cos\theta=8\), so \(3\sqrt{x^2+y^2}+2x=24\). Squaring and simplifying gives \(5x^2+9y^2+96x-576=0\). 7. For the circle, \(r^2=12r\cos\theta\), so \(x^2+y^2=12x\), or \((x-6)^2+y^2=36\). 8. Substituting \((3,\pm3\sqrt{3})\) satisfies both Cartesian equations. The pole is on the circle but not on the ellipse, so there are no additional intersections.

a) \(2c^2+3c-2=0\); valid polar intersections: \(\left(6,\frac{\pi}{3}\right)\), \(\left(6,-\frac{\pi}{3}\right)\) b) \((3,3\sqrt{3})\) and \((3,-3\sqrt{3})\) c) Ellipse: \(5x^2+9y^2+96x-576=0\); circle: \((x-6)^2+y^2=36\). Both Cartesian points satisfy both equations.

- Combine the vertical-line condition with the polar relation among \(x\), \(r\), and the direction angle. - Solve for the cosine of the direction angle before finding the radius. - Use symmetry to obtain both endpoints and then compute their vertical separation.

1. On the line \(x=2\), the polar relation \(x=r\cos\theta\) gives \(r\cos\theta=2\). 2. Substitute \(r=\frac{8}{1+\cos\theta}\): \(\frac{8\cos\theta}{1+\cos\theta}=2\). 3. Solving gives \(6\cos\theta=2\), so \(\cos\theta=\frac13\). 4. Therefore the two direction angles are \(\theta=\pm\arccos\left(\frac13\right)\), and \(r=\frac{8}{1+1/3}=6\). 5. The Cartesian \(y\)-coordinates are \(y=r\sin\theta=\pm6\sqrt{1-\frac19}=\pm4\sqrt2\). 6. The endpoints are \((2, 4\sqrt2)\) and \((2, -4\sqrt2)\), so the chord length is \(8\sqrt2\).

Polar points: \(\left(6, \arccos\left(\frac13\right)\right)\) and \(\left(6, -\arccos\left(\frac13\right)\right)\) Cartesian points: \((2, 4\sqrt2)\), \((2, -4\sqrt2)\) Chord length: \(8\sqrt2\)

- Convert each calibrated radius shown in the diagram into an equation for its reciprocal. - Use the two opposite horizontal measurements to determine the constant and cosine coefficients. - Use the vertical calibrated measurement to determine the sine coefficient. - Evaluate the completed model in the direction of the dashed reported measurement.

1. From \(r(0)=4\), \(A+B=\frac{1}{4}\). 2. From \(r(\pi)=4\), \(A-B=\frac{1}{4}\). 3. Solving gives \(A=\frac{1}{4}\) and \(B=0\). 4. From \(r\left(\frac{\pi}{2}\right)=\frac{12}{5}\), \(A+C=\frac{5}{12}\), so \(C=\frac{1}{6}\). 5. Thus \(\frac{1}{r}=\frac{1}{4}+\frac{1}{6}\sin\theta\), or \(r=\frac{4}{1+\frac{2}{3}\sin\theta}\). 6. Comparing with \(r=\frac{ed}{1+e\sin\theta}\), the eccentricity is \(e=\frac{2}{3}\). Since \(e<1\), the conic is an ellipse. 7. Also \(ed=4\), so \(d=\frac{4}{2/3}=6\). Therefore the associated directrix is \(y=6\). 8. At \(\theta=\frac{3\pi}{2}\), the model gives \(r=\frac{4}{1-\frac{2}{3}}=12\). 9. Therefore the reported value \(10\) is inconsistent, and the corrected value is \(12\).

Model: \(r=\frac{4}{1+\frac{2}{3}\sin\theta}\) Classification: ellipse Directrix: \(y=6\) The fourth report is inconsistent. Corrected value: \(r\left(\frac{3\pi}{2}\right)=12\)

- Write the radial distance for the opposite direction by changing the sign of the cosine term. - Multiply the two expressions before comparing directions. - The extreme values depend only on the possible range of the squared cosine.

1. The two opposite-ray distances are \(r_1=\frac{9}{1+\frac45\cos\theta}\) and \(r_2=\frac{9}{1-\frac45\cos\theta}\). 2. Their product is \(r_1r_2=\frac{81}{1-\frac{16}{25}\cos^2\theta}\). 3. The denominator is largest when \(\cos^2\theta=0\), giving the minimum product \(r_1r_2=81\). 4. This occurs for chords perpendicular to the polar axis, with \(\theta=\frac{\pi}{2}\) modulo \(\pi\). 5. The denominator is smallest when \(\cos^2\theta=1\), giving the maximum product \(r_1r_2=\frac{81}{9/25}=225\). 6. This occurs for chords along the polar axis, with \(\theta=0\) modulo \(\pi\).

Minimum product: \(81\), for chords perpendicular to the polar axis Maximum product: \(225\), for chords along the polar axis

- Read the eccentricity from the normalized polar denominator before converting coordinates. - Use \(r\sin\theta=y\) and \(r^2=x^2+y^2\). - Complete the square, then use \(c=ea\) to locate the second focus.

1. The equation has the focus-directrix form \(r=\frac{ed}{1+e\sin\theta}\), with eccentricity \(e=\frac25<1\). Therefore the conic is an ellipse. 2. Since \(ed=7\), \(d=\frac{7}{2/5}=\frac{35}{2}\). The associated directrix is \(y=\frac{35}{2}\). 3. Multiply the polar equation by its denominator: \(r+\frac25r\sin\theta=7\). Using \(r\sin\theta=y\), this becomes \(5r+2y=35\). 4. Substitute \(r=\sqrt{x^2+y^2}\) and square: \(25(x^2+y^2)=(35-2y)^2\). 5. Simplifying gives \(25x^2+21y^2+140y-1225=0\). 6. Complete the square in \(y\): \(25x^2+21\left(y+\frac{10}{3}\right)^2=\frac{4375}{3}\). 7. The Cartesian standard form is \(\frac{x^2}{175/3}+\frac{\left(y+\frac{10}{3}\right)^2}{625/9}=1\). 8. The center is \(\left(0,-\frac{10}{3}\right)\). The semi-major axis is \(a=\frac{25}{3}\), and the focal distance is \(c=ea=\frac{10}{3}\). 9. The vertices are \((0,5)\) and \(\left(0,-\frac{35}{3}\right)\). The foci are \((0,0)\) and \(\left(0,-\frac{20}{3}\right)\).

a) Ellipse; associated directrix \(y=\frac{35}{2}\) b) \(\frac{x^2}{175/3}+\frac{\left(y+\frac{10}{3}\right)^2}{625/9}=1\) c) Center: \(\left(0,-\frac{10}{3}\right)\); vertices: \((0,5)\) and \(\left(0,-\frac{35}{3}\right)\); foci: \((0,0)\) and \(\left(0,-\frac{20}{3}\right)\)

- Substitute the point whose angle makes the cosine term vanish first. - Use the second point to separate the eccentricity from the directrix distance. - In the Cartesian conversion, replace \(r\cos\theta\) with \(x\) before squaring.

1. For a right-hand directrix \(x=d\), the focus-directrix form is \(r=\frac{ed}{1+e\cos\theta}\). 2. At \(\theta=\frac{\pi}{2}\), \(\cos\theta=0\), so \(ed=6\). 3. At \(\theta=0\), \(4=\frac{6}{1+e}\). Thus \(1+e=\frac{3}{2}\), so \(e=\frac{1}{2}\) and \(d=12\). 4. The polar equation is \(r=\frac{6}{1+\frac{1}{2}\cos\theta}\). 5. Since \(r+\frac{1}{2}x=6\), squaring gives \(x^2+y^2=(6-\frac{x}{2})^2\), which simplifies to \(3x^2+4y^2+24x-144=0\). 6. Completing the square gives \(\frac{(x+4)^2}{64}+\frac{y^2}{48}=1\). 7. Here \(a^2=64\), \(b^2=48\), and \(c^2=16\). The foci are \((0, 0)\) and \((-8, 0)\).

Eccentricity: \(e=\frac{1}{2}\) Directrix: \(x=12\) Polar equation: \(r=\frac{6}{1+\frac{1}{2}\cos\theta}\) Cartesian form: \(\frac{(x+4)^2}{64}+\frac{y^2}{48}=1\) Foci: \((0, 0)\) and \((-8, 0)\)

- Recover the semi-major axis from eccentricity and semilatus rectum. - Use focal symmetry to locate the second focus. - Changing to the opposite focus reverses which side contains the associated directrix.

1. The eccentricity is \(e=\frac{1}{2}\), and the semilatus rectum is \(\ell=6\). 2. For an ellipse, \(\ell=a(1-e^2)\). Thus \(a=\frac{6}{1-1/4}=8\), and \(c=ea=4\). 3. The plus cosine form places the center to the left of the current focus, at \((-4, 0)\). Therefore the other focus is \((-8, 0)\). 4. Relative to the new pole, the associated directrix is \(12\) units to the left. 5. A left-hand directrix changes the denominator sign, while \(ed=\frac{1}{2}(12)=6\) remains the same. 6. The new polar equation is \(r_{\text{new}}=\frac{6}{1-\frac{1}{2}\cos\theta}\).

The new pole is at the original-coordinate point \((-8, 0)\). New polar equation: \(r_{\text{new}}=\frac{6}{1-\frac{1}{2}\cos\theta}\)

- Evaluate a standard polar-conic form on the two opposite axial directions. - Solve the two distance equations for the semilatus rectum and eccentricity. - Convert to Cartesian form only after the polar parameters are known.

1. Write the ellipse as \(r=\frac{\ell}{1+e\cos\theta}\), where \(\ell\) is the semilatus rectum. 2. The axial distances give \(\frac{\ell}{1+e}=4\) and \(\frac{\ell}{1-e}=12\). 3. Solving gives \(e=\frac{1}{2}\) and \(\ell=6\). 4. The polar equation is \(r=\frac{6}{1+\frac{1}{2}\cos\theta}\). 5. Since \(\ell=ed\), the associated directrix is \(x=d=12\). 6. From \(r+\frac{1}{2}x=6\), squaring gives \(x^2+y^2=\left(6-\frac{x}{2}\right)^2\). 7. Completing the square yields \(\frac{(x+4)^2}{64}+\frac{y^2}{48}=1\).

Eccentricity: \(e=\frac{1}{2}\) Semilatus rectum: \(\ell=6\) Polar equation: \(r=\frac{6}{1+\frac{1}{2}\cos\theta}\) Associated directrix: \(x=12\) Cartesian equation: \(\frac{(x+4)^2}{64}+\frac{y^2}{48}=1\)

- Replace \(r\sin\theta\) with y and isolate \(r\). - Square only after expressing the equation in Cartesian quantities. - Complete the square before reading the hyperbola's geometry.

1. Rewrite the equation as \(r-\frac{5}{3}r\sin\theta=4\), so \(r=4+\frac{5}{3}y\). 2. Square using \(r^2=x^2+y^2\): \(x^2+y^2=\left(4+\frac{5}{3}y\right)^2\). 3. Simplifying and completing the square gives \(16\left(y+\frac{15}{4}\right)^2-9x^2=81\). 4. The standard form is \(\frac{\left(y+\frac{15}{4}\right)^2}{81/16}-\frac{x^2}{9}=1\). 5. Thus, the center is \(\left(0, -\frac{15}{4}\right)\), with \(a=\frac{9}{4}\), \(b=3\), and \(c=\sqrt{a^2+b^2}=\frac{15}{4}\). 6. The vertices are \(\left(0, -\frac{3}{2}\right)\) and \((0, -6)\). The foci are \((0, 0)\) and \(\left(0, -\frac{15}{2}\right)\). 7. The asymptotes are \(y+\frac{15}{4}=\pm\frac{3}{4}x\).

Standard form: \(\frac{\left(y+\frac{15}{4}\right)^2}{81/16}-\frac{x^2}{9}=1\) Center: \(\left(0, -\frac{15}{4}\right)\) Vertices: \(\left(0, -\frac{3}{2}\right)\), \((0, -6)\) Foci: \((0, 0)\), \(\left(0, -\frac{15}{2}\right)\) Asymptotes: \(y+\frac{15}{4}=\pm\frac{3}{4}x\)

- Identify which directrix corresponds to the focus at the pole. - Express the point-to-directrix distance using \(r\sin\theta\). - Apply the focus-directrix ratio before isolating \(r\).

1. The hyperbola has center \((0, 5)\), \(a=3\), \(b=4\), and \(c=5\). 2. Its eccentricity is \(e=\frac{5}{3}\). The directrix associated with the focus at the pole is \(y=5-\frac{a^2}{c}=5-\frac{9}{5}=\frac{16}{5}\). 3. For a point \((r, \theta)\), the distance to that directrix is \(\frac{16}{5}-r\sin\theta\) on the branch associated with the pole. 4. The focus-directrix condition is \(r=\frac{5}{3}\left(\frac{16}{5}-r\sin\theta\right)\). 5. Simplifying gives \(r\left(1+\frac{5}{3}\sin\theta\right)=\frac{16}{3}\). 6. Therefore, \(r=\frac{16}{3+5\sin\theta}\).

\(r=\frac{16}{3+5\sin\theta}\)

- Isolate \(r\) after replacing \(r\cos\theta\) with \(x\). - Complete the square while keeping \(e\) symbolic. - Compare the two denominators and then examine the special value \(e=0\).

1. Rewrite the equation as \(r+ex=6\), so \(r=6-ex\). 2. Square using \(r^2=x^2+y^2\): \((1-e^2)x^2+y^2+12ex-36=0\). 3. Let \(A=1-e^2\). Completing the square gives \(A\left(x+\frac{6e}{A}\right)^2+y^2=\frac{36}{A}\). 4. Therefore, \(\frac{\left(x+\frac{6e}{1-e^2}\right)^2}{\frac{36}{(1-e^2)^2}}+\frac{y^2}{\frac{36}{1-e^2}}=1\). 5. The center is \(\left(-\frac{6e}{1-e^2}, 0\right)\), the semi-major axis is \(\frac{6}{1-e^2}\), and the semi-minor axis is \(\frac{6}{\sqrt{1-e^2}}\). 6. When \(e=0\), the center is the pole and both semi-axes equal \(6\), so the conic is the circle \(r=6\), or \(x^2+y^2=36\).

Center: \(\left(-\frac{6e}{1-e^2}, 0\right)\) Semi-major axis: \(\frac{6}{1-e^2}\) Semi-minor axis: \(\frac{6}{\sqrt{1-e^2}}\) At \(e=0\): the circle \(x^2+y^2=36\)

- Convert the polar ellipse to Cartesian form first. - Intersect it with a general line through the pole and use the sum of the two roots. - Eliminate the line's slope from the midpoint coordinates.

1. Since \(r+\frac23x=9\), converting to Cartesian form gives \(5x^2+9y^2+108x-729=0\). 2. A nonvertical line through the pole has equation \(y=mx\). Substitution gives \((5+9m^2)x^2+108x-729=0\). 3. If \(x_1\) and \(x_2\) are the endpoint x-coordinates, then \(x_1+x_2=-\frac{108}{5+9m^2}\). The midpoint therefore has \(x=-\frac{54}{5+9m^2}\) and \(y=mx\). 4. Eliminating \(m=\frac{y}{x}\) gives \(5x^2+9y^2=-54x\). 5. Completing the square gives \(\frac{\left(x+\frac{27}{5}\right)^2}{729/25}+\frac{y^2}{81/5}=1\). The vertical focal chord has midpoint \((0,0)\), which also lies on this ellipse.

The midpoint locus is \(\frac{\left(x+\frac{27}{5}\right)^2}{729/25}+\frac{y^2}{81/5}=1\).

- Express the original Cartesian coordinates in terms of the new pole, direction, and radial distance. - The substitution produces one solution corresponding to the pole itself and another for the second intersection. - Compare the structure of the result with a polar equation whose pole is a focus.

1. Substitute \(x=10-r\cos\theta\) and \(y=r\sin\theta\) into the ellipse equation. 2. After expanding and canceling the constant term, the equation becomes \(-\frac{r\cos\theta}{5}+r^2\left(\frac{\cos^2\theta}{100}+\frac{\sin^2\theta}{75}\right)=0\). 3. One solution is \(r=0\), which is the pole \(V\). The other algebraic solution is \(r=\frac{60\cos\theta}{3\cos^2\theta+4\sin^2\theta}\). 4. The denominator is always positive, so this second solution is a positive ray distance exactly when \(\cos\theta>0\). If \(\cos\theta=0\), the ray is tangent at \(V\) and has no second point. If \(\cos\theta<0\), the nonzero algebraic radius is negative and represents a point on the opposite ray. 5. At \(\theta=\frac{\pi}{3}\), \(r=\frac{60(1/2)}{3(1/4)+4(3/4)}=8\). 6. Convert back to Cartesian coordinates: \(x=10-8\cdot\frac12=6\), \(y=8\cdot\frac{\sqrt3}{2}=4\sqrt3\). 7. The standard focus-directrix polar form places a focus at the pole and has a denominator linear in \(\cos\theta\) or \(\sin\theta\). Here the pole is a vertex, and the denominator contains squared trigonometric terms.

a) For \(\cos\theta>0\), the second intersection is at \(r=\frac{60\cos\theta}{3\cos^2\theta+4\sin^2\theta}\). There is no second point on the ray when \(\cos\theta\le0\). b) \(r=8\), giving \((6,4\sqrt3)\) c) The pole is a vertex rather than a focus, so the equation is not a focus-directrix polar conic form.

- Translate each displayed radial measurement into an equation for the reciprocal-radius model. - Combine the cosine and sine terms into one shifted cosine after finding their coefficients. - Compare the normalized denominator with the focus-directrix polar form to read the eccentricity and directrix.

1. From the displayed measurement \(r(0)=\frac{5}{2}\), \(A+B=\frac{2}{5}\). 2. From \(r(\pi)=10\), \(A-B=\frac{1}{10}\). 3. Solving gives \(A=\frac{1}{4}\) and \(B=\frac{3}{20}\). 4. From \(r\left(\frac{\pi}{2}\right)=\frac{20}{9}\), \(A+C=\frac{9}{20}\), so \(C=\frac{1}{5}\). 5. Therefore, \(\frac{1}{r}=\frac{5+3\cos\theta+4\sin\theta}{20}\), so \(r=\frac{20}{5+3\cos\theta+4\sin\theta}\). 6. Since \(3\cos\theta+4\sin\theta=5\cos(\theta-\phi)\), where \(\cos\phi=\frac{3}{5}\) and \(\sin\phi=\frac{4}{5}\), the equation is \(r=\frac{4}{1+\cos(\theta-\phi)}\). 7. The eccentricity is \(1\), so the conic is a parabola. Its directrix is \(x\cos\phi+y\sin\phi=4\), or \(3x+4y=20\).

Polar equation: \(r=\frac{20}{5+3\cos\theta+4\sin\theta}\) Classification: parabola Directrix: \(3x+4y=20\)