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.

- 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 \(x=p\), the focus–directrix form is \(r=\frac{ep}{1+e\cos\theta}\). 2. Here \(ep=\frac{2}{3}(9)=6\), 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) \(r(0)=\frac{18}{5}\), \(r(\pi)=18\)

- 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 \(\theta=\frac{7\pi}{6}\)

- Evaluate the polar equation in the two directions along its symmetry axis. - Use those axial points to find the center and semi-major axis geometrically. - Use eccentricity to locate the second focus and remaining semi-axis.

1. Rewrite the equation as \(r=\frac{4}{1+\frac{1}{3}\cos\theta}\), so the eccentricity is \(e=\frac{1}{3}\). 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, \((-\frac{3}{2}, 0)\), 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 \(\frac{(x+\frac{3}{2})^2}{81/4}+\frac{y^2}{18}=1\).

Center: \((-\frac{3}{2}, 0)\) Vertices: \((3, 0)\), \((-6, 0)\) Foci: \((0, 0)\), \((-3, 0)\) Equation: \(\frac{(x+\frac{3}{2})^2}{81/4}+\frac{y^2}{18}=1\)

- 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: \(\cos\left(\phi+\frac{\pi}{3}\right)=\frac12\cos\phi-\frac{\sqrt3}{2}\sin\phi\). 3. The new polar equation is \(r=\frac{8}{1+\frac16\cos\phi-\frac{\sqrt3}{6}\sin\phi}\). 4. The original equation has eccentricity \(\frac13\) and semilatus rectum \(8\), so its associated directrix is \(x=24\). 5. The coordinate relation is \(x=X\cos60^\circ-Y\sin60^\circ=\frac12X-\frac{\sqrt3}{2}Y\). 6. Therefore the directrix in the new coordinates is \(\frac12X-\frac{\sqrt3}{2}Y=24\), or \(X-\sqrt3Y=48\).

New polar equation: \(r=\frac{8}{1+\frac16\cos\phi-\frac{\sqrt3}{6}\sin\phi}\) Directrix in the new coordinates: \(X-\sqrt3Y=48\)

- 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 \(\left(\frac{16}{5}, -\frac{12}{5}\right)\)

- Normalize the denominator to read the semilatus rectum and eccentricity. - Relate the semilatus rectum to the ellipse's semi-major axis. - Use eccentricity to recover the other semi-axis before finding the area.

1. Normalize the equation: \(r=\frac{6}{1+\frac{1}{3}\cos\theta}\). 2. Thus, the semilatus rectum is \(\ell=6\) and the eccentricity is \(e=\frac{1}{3}\). 3. For an ellipse, \(\ell=a(1-e^2)\), so \(a=\frac{6}{1-1/9}=\frac{27}{4}\). 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 \(\pi ab=\pi\left(\frac{27}{4}\right)\left(\frac{9\sqrt{2}}{2}\right)=\frac{243\sqrt{2}\pi}{8}\).

Eccentricity: \(\frac{1}{3}\) Semi-major axis: \(\frac{27}{4}\) Semi-minor axis: \(\frac{9\sqrt{2}}{2}\) Area: \(\frac{243\sqrt{2}\pi}{8}\)

- 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.

- At a non-pole intersection, the two curves have the same radius for the same direction angle. - After solving for \(\cos\theta\), reject values outside \([-1,1]\) and recover all corresponding angles. - Use \(r^2=x^2+y^2\) and \(r\cos\theta=x\) to convert each equation to Cartesian form.

1. At a common non-pole point, equate the radial expressions: \(\frac{8}{1+\frac23\cos\theta}=12\cos\theta\). 2. Let \(c=\cos\theta\). Then \(8=12c+8c^2\), so \(2c^2+3c-2=0\). 3. The roots are \(c=\frac12\) and \(c=-2\). Only \(c=\frac12\) is possible for a cosine. 4. Thus \(\theta=\pm\frac{\pi}{3}\), and \(r=12\left(\frac12\right)=6\). 5. The Cartesian intersection points are \((r\cos\theta,r\sin\theta)=(3,\pm3\sqrt3)\). 6. For the ellipse, \(r+\frac23r\cos\theta=8\), 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 \((3,\pm3\sqrt3)\) satisfies both Cartesian equations. The pole is on the circle but not on the ellipse, so there are no additional intersections.

a) \((3,3\sqrt3)\) and \((3,-3\sqrt3)\) b) Ellipse: \(5x^2+9y^2+96x-576=0\); circle: \((x-6)^2+y^2=36\). Both intersection 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 into an equation for its reciprocal. - Use the opposite horizontal directions to determine the constant and cosine coefficients first. - Evaluate the completed model at the reported fourth angle before accepting the measurement.

1. From \(r(0)=4\), \(A+B=\frac14\). 2. From \(r(\pi)=4\), \(A-B=\frac14\). 3. Solving gives \(A=\frac14\) and \(B=0\). 4. From \(r\left(\frac{\pi}{2}\right)=\frac{12}{5}\), \(A+C=\frac{5}{12}\), so \(C=\frac16\). 5. Thus \(\frac1r=\frac14+\frac16\sin\theta\), or \(r=\frac4{1+\frac23\sin\theta}\). 6. The eccentricity is \(\frac23\), so the conic is an ellipse. Its semilatus rectum is \(4\), so the associated directrix is \(y=\frac{4}{2/3}=6\). 7. At \(\theta=\frac{3\pi}{2}\), \(r=\frac4{1-\frac23}=12\). Therefore the reported value \(10\) is inconsistent.

Model: \(r=\frac4{1+\frac23\sin\theta}\) 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 \(r'=\frac{6}{1-\frac{1}{2}\cos\theta}\).

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

- 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.

- Substitute each measured angle into 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 \(r(0)=\frac52\), \(A+B=\frac25\). 2. From \(r(\pi)=10\), \(A-B=\frac1{10}\). 3. Solving gives \(A=\frac14\) and \(B=\frac3{20}\). 4. From \(r\left(\frac{\pi}{2}\right)=\frac{20}{9}\), \(A+C=\frac9{20}\), so \(C=\frac15\). 5. Therefore \(\frac1r=\frac{5+3\cos\theta+4\sin\theta}{20}\), so \(r=\frac{20}{5+3\cos\theta+4\sin\theta}\). 6. Since \(3\cos\theta+4\sin\theta=5\cos(\theta-\phi)\), where \(\cos\phi=\frac35\) and \(\sin\phi=\frac45\), the equation is \(r=\frac4{1+\cos(\theta-\phi)}\). 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: \(3x+4y=20\)