Free math worksheets

- Read the center from the horizontal and vertical shifts. - The positive squared term gives the transverse-axis direction. - For a horizontal hyperbola, the asymptote slope magnitude is \(\frac{b}{a}\).

1. The center is \((-1, 3)\). 2. The positive x-term shows that the transverse axis is horizontal. Since \(a=7\), the vertices are \((-8, 3)\) and \((6, 3)\). 3. Here \(b=4\), so the asymptote slope magnitude is \(\frac{b}{a}=\frac{4}{7}\). 4. The asymptotes are \(y-3=\frac{4}{7}(x+1)\) and \(y-3=-\frac{4}{7}(x+1)\).

Center: \((-1, 3)\) Vertices: \((-8, 3)\), \((6, 3)\) Asymptotes: \(y-3=\frac{4}{7}(x+1)\) and \(y-3=-\frac{4}{7}(x+1)\)

- Read the center from the shifts in standard form. - The positive squared term identifies the transverse-axis direction. - For a vertical hyperbola, the asymptote slope magnitude is \(\frac{a}{b}\).

1. The center is \((-2, 1)\). 2. The positive y-term shows that the transverse axis is vertical. Since \(a=5\), the vertices are \((-2, 6)\) and \((-2, -4)\). 3. Here \(b=3\), so the asymptote slope magnitude for a vertical hyperbola is \(\frac{a}{b}=\frac{5}{3}\). 4. The asymptotes are \(y-1=\frac{5}{3}(x+2)\) and \(y-1=-\frac{5}{3}(x+2)\).

Center: \((-2, 1)\) Vertices: \((-2, 6)\), \((-2, -4)\) Asymptotes: \(y-1=\frac{5}{3}(x+2)\) and \(y-1=-\frac{5}{3}(x+2)\)

- A conjugate hyperbola reverses which squared term is positive. - Compare the guide rectangles of the two standard forms. - Test possible common points by combining the two equations.

1. Interchanging the positive and negative squared terms gives the conjugate hyperbola \(\frac{y^2}{4}-\frac{x^2}{25}=1\). 2. The vertices of \(H\) are \((\pm5, 0)\). The vertices of the conjugate hyperbola are \((0, \pm2)\). 3. Both hyperbolas use the same asymptote rectangle, so their asymptotes are \(y=\pm\frac{2}{5}x\). 4. If a point satisfied both equations, adding them would give \(0=2\), which is impossible. 5. Therefore the two conjugate hyperbolas do not intersect.

a) \(\frac{y^2}{4}-\frac{x^2}{25}=1\) b) Vertices of \(H\): \((\pm5, 0)\); vertices of the conjugate: \((0, \pm2)\); asymptotes: \(y=\pm\frac{2}{5}x\) c) They do not intersect.

- Read \(a\) from the center-to-vertex distance. - Use the marked point to determine the slope of the upper-right asymptote. - For a horizontal hyperbola, the asymptote slope magnitude is \(\frac{b}{a}\).

1. The vertices are \((\pm3, 0)\), so \(a=3\) and \(a^2=9\). 2. The upper-right asymptote passes through the origin and \(A=(3, 2)\), so its slope is \(\frac{2}{3}\). 3. For a horizontal hyperbola, the asymptote slope is \(\frac{b}{a}\). Thus \(\frac{b}{3}=\frac{2}{3}\), so \(b=2\) and \(b^2=4\). 4. The standard form is \(\frac{x^2}{9}-\frac{y^2}{4}=1\). 5. The asymptotes are \(y=\frac{2}{3}x\) and \(y=-\frac{2}{3}x\).

Standard form: \(\frac{x^2}{9}-\frac{y^2}{4}=1\) Asymptotes: \(y=\frac{2}{3}x\) and \(y=-\frac{2}{3}x\)

- Use the center and vertices to determine \(a\). - For a vertical hyperbola, the asymptote slope magnitude is \(\frac{a}{b}\). - Place the positive term under the variable aligned with the transverse axis.

1. The center-to-vertex distance is \(a=4\), so \(a^2=16\). 2. For a vertical hyperbola, the asymptote slope magnitude is \(\frac{a}{b}\). 3. Since \(\frac{a}{b}=2\) and \(a=4\), it follows that \(b=2\), so \(b^2=4\). 4. The standard form is \(\frac{(y+3)^2}{16}-\frac{(x-2)^2}{4}=1\).

\(\frac{(y+3)^2}{16}-\frac{(x-2)^2}{4}=1\)

- Keep the negative sign attached to the complete y-group while completing the square. - The positive squared term identifies the transverse-axis direction. - Use the square roots of the denominators to form the asymptote slopes.

1. Group the variable terms: \(25(x^2-8x)-16(y^2+4y)-64=0\). 2. Complete both squares: \(25((x-4)^2-16)-16((y+2)^2-4)-64=0\). 3. Simplify to \(25(x-4)^2-16(y+2)^2=400\). 4. Divide by \(400\): \(\frac{(x-4)^2}{16}-\frac{(y+2)^2}{25}=1\). 5. The graph is a horizontal hyperbola centered at \((4, -2)\), with vertices \((0, -2)\) and \((8, -2)\). 6. Its asymptotes are \(y+2=\pm\frac{5}{4}(x-4)\).

The conic is a horizontal hyperbola with standard form \(\frac{(x-4)^2}{16}-\frac{(y+2)^2}{25}=1\). Center: \((4, -2)\) Vertices: \((0, -2)\), \((8, -2)\) Asymptotes: \(y+2=\pm\frac{5}{4}(x-4)\)

- Use the center-to-vertex distance to determine \(a\). - For a horizontal hyperbola, compare each asymptote slope with \(\pm\frac{b}{a}\). - Translate the horizontal-hyperbola standard form to the given center.

1. The vertices are \(5\) units to the left and right of the center, so \(a=5\). 2. For a horizontal hyperbola, the asymptotes have slopes \(\pm\frac{b}{a}\). 3. Since the given slopes are \(\pm\frac{3}{5}\), \(b=3\). 4. The standard form is \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\). 5. Substituting \((h,k)=(3,1)\), \(a=5\), and \(b=3\) gives \(\frac{(x-3)^2}{25}-\frac{(y-1)^2}{9}=1\).

\(\frac{(x-3)^2}{25}-\frac{(y-1)^2}{9}=1\)

- The focus locations determine the transverse-axis direction and the value of \(c\). - For a horizontal hyperbola, the asymptote slopes are \(\pm\frac{b}{a}\). - Use the hyperbola relation \(c^2=a^2+b^2\), which differs from the ellipse relation.

1. The foci lie on the x-axis, so the hyperbola opens left and right and has form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). 2. The focal distance is \(c=13\), and the asymptote slopes give \(\frac{b}{a}=\frac{12}{5}\). 3. Write \(a=5k\) and \(b=12k\). 4. For a hyperbola, \(c^2=a^2+b^2\), so \(169=25k^2+144k^2=169k^2\). 5. Since lengths are positive, \(k=1\), giving \(a=5\) and \(b=12\). 6. The equation is \(\frac{x^2}{25}-\frac{y^2}{144}=1\), and the vertices are \((\pm5,0)\).

Equation: \(\frac{x^2}{25}-\frac{y^2}{144}=1\) Vertices: \((-5, 0)\) and \((5, 0)\)

- Read \(a\) from the center-to-vertex distance and \(c\) from the center-to-focus distance. - For a hyperbola, use \(c^2=a^2+b^2\), not the ellipse relation. - Translate the asymptotes \(y=\pm\frac{b}{a}x\) to the given center.

1. The center-to-vertex distance is \(a=5\), and the center-to-focus distance is \(c=13\). 2. For a hyperbola, \(c^2=a^2+b^2\), so \(b^2=169-25=144\). 3. A horizontal hyperbola centered at \((h,k)\) has equation \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\). 4. Therefore, the equation is \(\frac{(x+3)^2}{25}-\frac{(y-4)^2}{144}=1\). 5. Its asymptotes are \(y-4=\pm\frac{12}{5}(x+3)\).

Equation: \(\frac{(x+3)^2}{25}-\frac{(y-4)^2}{144}=1\) Asymptotes: \(y-4=\pm\frac{12}{5}(x+3)\)

- Use the segment between the foci as the triangle's base. - The perpendicular height to that horizontal base is \(|y|\). - Substitute the resulting y-coordinate into the hyperbola and include all sign combinations.

1. The focal distance is \(c=\sqrt{49+576}=25\), so the foci are \((-25,0)\) and \((25,0)\). 2. The focal segment has length \(50\). Using it as the triangle base, the area is \(\frac{1}{2}(50)|y|=25|y|\). 3. Setting \(25|y|=600\) gives \(|y|=24\). 4. Substitute \(y^2=576\) into the hyperbola: \(\frac{x^2}{49}-1=1\). 5. Thus, \(x^2=98\), so \(x=\pm7\sqrt{2}\). 6. Both signs of \(x\) and both signs of \(y\) are possible.

The four points are \((7\sqrt{2}, 24)\), \((7\sqrt{2}, -24)\), \((-7\sqrt{2}, 24)\), and \((-7\sqrt{2}, -24)\).

- Translate the coordinates so the two asymptotes become the coordinate axes. - Use the product form \(UV=k\) and substitute the known point to determine \(k\). - The sign of \(k\) determines whether the shifted coordinates have matching or opposite signs on the branches.

1. Shift to \(U=x-1\) and \(V=y+2\), so the asymptotes become \(U=0\) and \(V=0\). 2. A rectangular hyperbola with these asymptotes has equation \(UV=k\). 3. The point \((4,0)\) gives \(U=3\) and \(V=2\), so \(k=6\). 4. Therefore, the equation is \((x-1)(y+2)=6\). 5. The center is the intersection of the asymptotes, \((1,-2)\). 6. Since the product is positive, the branches lie where \(x-1\) and \(y+2\) have the same sign: upper right and lower left relative to the center.

Equation: \((x-1)(y+2)=6\) Center: \((1, -2)\) Branches: upper right and lower left relative to the center

- Find \(c\), eccentricity, and the two directrix locations from the standard equation. - Remove the absolute values separately for points on the right and left branches. - Compare the simplified distance differences and interpret their signs.

1. Here \(a=12\), \(b=5\), and \(c=\sqrt{144+25}=13\), so \(e=\frac{13}{12}\). 2. The directrices are \(x=\pm\frac{a}{e}=\pm\frac{144}{13}\). 3. On the right branch, \(x\ge12\). Thus, \(d_L=x+\frac{144}{13}\) and \(d_R=x-\frac{144}{13}\), so \(d_L-d_R=\frac{288}{13}\). 4. On the left branch, \(x\le-12\). Thus, \(d_L=-x-\frac{144}{13}\) and \(d_R=-x+\frac{144}{13}\), so \(d_L-d_R=-\frac{288}{13}\). 5. Therefore, the sign of the constant difference identifies the branch.

Right branch: \(d_L-d_R=\frac{288}{13}\) Left branch: \(d_L-d_R=-\frac{288}{13}\) A positive difference identifies the right branch, and a negative difference identifies the left branch.

- Use the center-to-vertex distance to determine \(a\). - For a vertical hyperbola, compare the asymptote slopes with \(\pm\frac{a}{b}\). - Use \(c^2=a^2+b^2\) and place the foci along the vertical transverse axis.

1. The center-to-vertex distance is \(a=5\). 2. For a vertical hyperbola, the asymptote slopes are \(\pm\frac{a}{b}\). 3. Since \(\frac{a}{b}=\frac{5}{4}\), \(b=4\). 4. The equation is \(\frac{(y+1)^2}{25}-\frac{(x-2)^2}{16}=1\). 5. For a hyperbola, \(c^2=a^2+b^2=25+16=41\), so \(c=\sqrt{41}\). 6. The foci are \((2,-1\pm\sqrt{41})\).

Equation: \(\frac{(y+1)^2}{25}-\frac{(x-2)^2}{16}=1\) Foci: \((2, -1-\sqrt{41})\) and \((2, -1+\sqrt{41})\)

- Read each asymptote slope from the ratio of the two semi-axis parameters. - Compare the two left-hand expressions before solving a system. - Conjugate hyperbolas exchange which squared term is positive while retaining the same asymptotes.

1. For \(H_1\), setting the right side informally to zero in the translated homogeneous part gives asymptotes \(y=\pm\frac{2}{3}x\). 2. Reversing the positive and negative squared terms for \(H_2\) produces the same two asymptote lines. 3. The left side of the equation for \(H_2\) is the negative of the left side for \(H_1\). 4. A common point would require one expression to equal both \(1\) and \(-1\), which is impossible. 5. Thus, the hyperbolas do not intersect. They share a center and asymptotes, and their transverse axes are perpendicular.

a) Both hyperbolas have asymptotes \(y=\pm\frac{2}{3}x\). b) They have no finite intersection points. c) They are conjugate hyperbolas with the same center and asymptotes and perpendicular transverse axes.

- Use the vertex distance from the center to determine \(a\). - For a vertical hyperbola, the asymptote slopes are \(\pm\frac{a}{b}\). - Use the hyperbola relationship \(c^2=a^2+b^2\) to locate the foci.

1. The vertices are \(4\) units above and below the center, so \(a=4\) and \(a^2=16\). 2. For a vertical hyperbola, the asymptotes have slopes \(\pm\frac{a}{b}\). 3. Since \(\frac{a}{b}=\frac23\), \(\frac{4}{b}=\frac23\), so \(b=6\) and \(b^2=36\). 4. The standard equation is \(\frac{(y-5)^2}{16}-\frac{x^2}{36}=1\). 5. For a hyperbola, \(c^2=a^2+b^2=16+36=52\), so \(c=2\sqrt{13}\). 6. The foci lie on the vertical transverse axis at \((0, 5\pm2\sqrt{13})\).

Equation: \(\frac{(y-5)^2}{16}-\frac{x^2}{36}=1\) Foci: \((0, 5+2\sqrt{13})\) and \((0, 5-2\sqrt{13})\)

- Use the center-to-vertex distance to determine the transverse semi-axis. - For a horizontal hyperbola, compare the asymptote slopes with \(\pm\frac{b}{a}\). - Use \(c^2=a^2+b^2\) after finding both denominators.

1. The distance from the center \((2, -1)\) to the vertex \((8, -1)\) is \(a=6\), so \(a^2=36\). 2. For a horizontal hyperbola, the asymptote slopes are \(\pm\frac{b}{a}\). 3. Since \(\frac{b}{a}=\frac56\) and \(a=6\), it follows that \(b=5\) and \(b^2=25\). 4. The standard equation is \(\frac{(x-2)^2}{36}-\frac{(y+1)^2}{25}=1\). 5. The other vertex is \((2-6, -1)=(-4, -1)\). 6. For a hyperbola, \(c^2=a^2+b^2=36+25=61\), so \(c=\sqrt{61}\). 7. The foci are \((2\pm\sqrt{61}, -1)\).

Equation: \(\frac{(x-2)^2}{36}-\frac{(y+1)^2}{25}=1\) Other vertex: \((-4, -1)\) Foci: \((2+\sqrt{61}, -1)\) and \((2-\sqrt{61}, -1)\)

- First locate the foci of the hyperbola. - A latus rectum passes through a focus and is perpendicular to the transverse axis. - Use the four symmetric endpoints to determine the rectangle's dimensions.

1. The hyperbola has \(a=4\), \(b^2=20\), and \(c=\sqrt{a^2+b^2}=6\), so the foci are \((\pm6, 0)\). 2. Each latus rectum is perpendicular to the transverse axis and passes through a focus, so its supporting line is \(x=6\) or \(x=-6\). 3. Substitute \(x=6\) into the hyperbola: \(\frac{36}{16}-\frac{y^2}{20}=1\). 4. This gives \(\frac{y^2}{20}=\frac54\), so \(y=\pm5\). 5. The four endpoints are \((6,\pm5)\) and \((-6,\pm5)\). 6. Each latus rectum has length \(10\). 7. The rectangle has width \(12\) and height \(10\), so its area is \(120\).

Latus-rectum endpoints: \((6,\pm5)\) and \((-6,\pm5)\) Length of each latus rectum: \(10\) Rectangle area: \(120\)

- Read the center and semi-axis parameters directly from standard form. - Use \(c^2=a^2+b^2\) for the foci. - The asymptote slopes determine equal and opposite direction angles from the horizontal.

1. The hyperbola is horizontal with center \((1, -2)\), \(a=7\), and \(b=4\). 2. The vertices are \((1\pm7,-2)\), or \((8,-2)\) and \((-6,-2)\). 3. The focal distance satisfies \(c^2=a^2+b^2=49+16=65\), so \(c=\sqrt{65}\). 4. The foci are \((1\pm\sqrt{65},-2)\). 5. The asymptotes are \(y+2=\pm\frac47(x-1)\). 6. Each asymptote makes an angle \(\arctan\left(\frac47\right)\) with the horizontal axis. Therefore the acute angle between them is \(2\arctan\left(\frac47\right)\approx59.5^\circ\).

Center: \((1,-2)\) Vertices: \((8,-2)\) and \((-6,-2)\) Foci: \((1+\sqrt{65},-2)\) and \((1-\sqrt{65},-2)\) Asymptotes: \(y+2=\pm\frac47(x-1)\) Acute angle: \(2\arctan\left(\frac47\right)\approx59.5^\circ\)

- Start with the standard form that matches a vertical transverse axis. - Use the asymptote slope to relate the two denominator values. - Substitute the given point only after expressing both denominators with one common scale factor.

1. For a vertical hyperbola, the standard form is \(\frac{(y+1)^2}{a^2}-\frac{(x-2)^2}{b^2}=1\), and the asymptote slope is \(\frac{a}{b}\). 2. Since \(\frac{a}{b}=\frac{3}{2}\), let \(a^2=9k\) and \(b^2=4k\) for some positive \(k\). 3. Substituting \((6, 8)\) gives \(\frac{81}{9k}-\frac{16}{4k}=1\), so \(\frac{5}{k}=1\) and \(k=5\). 4. Thus \(a^2=45\) and \(b^2=20\), so the equation is \(\frac{(y+1)^2}{45}-\frac{(x-2)^2}{20}=1\). 5. Since \(a=3\sqrt{5}\), the vertices are \((2, -1+3\sqrt{5})\) and \((2, -1-3\sqrt{5})\).

The hyperbola is \(\frac{(y+1)^2}{45}-\frac{(x-2)^2}{20}=1\). Its vertices are \((2, -1+3\sqrt{5})\) and \((2, -1-3\sqrt{5})\).

- Use the center and vertices to determine the transverse-axis direction and one semi-axis length. - Find the slope of the given asymptote from the two points on it. - Relate that slope to the dimensions of the hyperbola's asymptote rectangle.

1. The vertices are vertical from the center, so the hyperbola has a vertical transverse axis. Their distance from the center gives \(a=5\). 2. The slope from the center to \((1, 8)\) is \(\frac{8-2}{1-(-3)}=\frac{3}{2}\). 3. For a vertical hyperbola, an asymptote slope has magnitude \(\frac{a}{b}\). Thus \(\frac{5}{b}=\frac{3}{2}\), so \(b=\frac{10}{3}\) and \(b^2=\frac{100}{9}\). 4. The standard form is \(\frac{(y-2)^2}{25}-\frac{(x+3)^2}{100/9}=1\). 5. The asymptotes are \(y-2=\frac{3}{2}(x+3)\) and \(y-2=-\frac{3}{2}(x+3)\).

The hyperbola is \(\frac{(y-2)^2}{25}-\frac{(x+3)^2}{100/9}=1\). Its asymptotes are \(y-2=\frac{3}{2}(x+3)\) and \(y-2=-\frac{3}{2}(x+3)\).

- Use eccentricity to relate the two semi-axis lengths. - Translate the rectangle’s area into a product involving those lengths. - The asymptote slopes come from the rectangle’s side-length ratio.

1. For a horizontal hyperbola, \(c^2=a^2+b^2\) and \(e=\frac{c}{a}=\frac{5}{4}\). 2. Thus \(\frac{b^2}{a^2}=e^2-1=\frac{25}{16}-1=\frac{9}{16}\), so \(b=\frac{3}{4}a\). 3. The rectangle has width \(2a\) and height \(2b\), so \(4ab=96\), or \(ab=24\). 4. Substituting \(b=\frac{3}{4}a\) gives \(\frac{3}{4}a^2=24\), so \(a^2=32\) and \(b^2=18\). 5. The equation is \(\frac{x^2}{32}-\frac{y^2}{18}=1\), and the asymptotes are \(y=\pm\frac{b}{a}x=\pm\frac{3}{4}x\).

\(\frac{x^2}{32}-\frac{y^2}{18}=1\) Asymptotes: \(y=\frac{3}{4}x\) and \(y=-\frac{3}{4}x\)

- Write the two asymptote slopes in terms of the semi-axis lengths. - Use the condition for two nonvertical lines to be perpendicular. - After finding both semi-axes, determine the focal distance.

1. The vertices give \(a=3\). 2. The asymptotes of \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) have slopes \(\pm\frac{b}{a}\). 3. Their slope product is \(-\frac{b^2}{a^2}\). Perpendicular lines have slope product \(-1\), so \(b=a=3\). 4. The equation is \(\frac{x^2}{9}-\frac{y^2}{9}=1\), and the asymptotes are \(y=x\) and \(y=-x\). 5. Since \(c^2=a^2+b^2=18\), \(c=3\sqrt{2}\). The foci are \((\pm3\sqrt{2}, 0)\), and \(e=\frac{c}{a}=\sqrt{2}\).

Equation: \(\frac{x^2}{9}-\frac{y^2}{9}=1\) Asymptotes: \(y=x\), \(y=-x\) Foci: \((3\sqrt{2}, 0)\), \((-3\sqrt{2}, 0)\) Eccentricity: \(\sqrt{2}\)

- Substitute the line through the center into the hyperbola. - The sign of the resulting coefficient of \(x^2\) controls whether real points exist. - Compare the boundary slopes with the hyperbola’s asymptotes.

1. Substitute \(y=mx\): \(x^2(\frac{1}{36}-\frac{m^2}{25})=1\). 2. If \(\frac{1}{36}-\frac{m^2}{25}>0\), there are two intersections. This occurs when \(|m|<\frac{5}{6}\). 3. If \(|m|=\frac{5}{6}\), the line is an asymptote and the equation becomes \(0=1\), so there is no finite intersection. 4. If \(|m|>\frac{5}{6}\), the coefficient is negative, so there is no real intersection. 5. For \(m=\frac{1}{2}\), \(x^2(\frac{1}{36}-\frac{1}{100})=1\), giving \(x^2=\frac{225}{4}\). 6. Thus the points are \((\frac{15}{2}, \frac{15}{4})\) and \((-\frac{15}{2}, -\frac{15}{4})\).

\(|m|<\frac{5}{6}\): two finite intersections \(|m|\ge\frac{5}{6}\): no finite intersections For \(m=\frac{1}{2}\): \((\frac{15}{2}, \frac{15}{4})\) and \((-\frac{15}{2}, -\frac{15}{4})\)

- Determine the focal distance from the two denominator values. - Pair the right branch with the focus and directrix on the right. - Compute the two distances separately before taking their ratio.

1. Here \(a=6\), \(b^2=13\), and \(c^2=a^2+b^2=49\), so \(c=7\). 2. The foci are \((\pm7, 0)\), and the eccentricity is \(e=\frac{c}{a}=\frac{7}{6}\). 3. The directrices are \(x=\pm\frac{a}{e}=\pm\frac{36}{7}\). 4. For the right branch, pair the right focus \((7, 0)\) with the right directrix \(x=\frac{36}{7}\). 5. The distance from \(P\) to the right focus is \(\frac{13}{6}\). 6. The perpendicular distance from \(P\) to the right directrix is \(7-\frac{36}{7}=\frac{13}{7}\). 7. Their ratio is \(\frac{13/6}{13/7}=\frac{7}{6}=e\).

Eccentricity: \(\frac{7}{6}\) Foci: \((7, 0)\), \((-7, 0)\) Directrices: \(x=\frac{36}{7}\), \(x=-\frac{36}{7}\) At \(P\), \(\frac{PF}{\text{distance to directrix}}=\frac{13/6}{13/7}=\frac{7}{6}\).

- Use the asymptote slopes to relate the two semi-axis lengths. - Express eccentricity from that ratio. - Use the directrix distance to determine the scale of the hyperbola.

1. For \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), the asymptote slope gives \(\frac{b}{a}=2\), so \(b=2a\). 2. Then \(c^2=a^2+b^2=5a^2\), so \(c=a\sqrt{5}\) and \(e=\sqrt{5}\). 3. The directrix distance is \(\frac{a}{e}=\frac{a}{\sqrt{5}}\). Setting this equal to \(3\) gives \(a=3\sqrt{5}\). 4. Thus \(a^2=45\), \(b^2=180\), and \(c=15\). 5. The equation is \(\frac{x^2}{45}-\frac{y^2}{180}=1\). The vertices are \((\pm3\sqrt{5}, 0)\), and the foci are \((\pm15, 0)\).

\(\frac{x^2}{45}-\frac{y^2}{180}=1\) Vertices: \((3\sqrt{5}, 0)\), \((-3\sqrt{5}, 0)\) Foci: \((15, 0)\), \((-15, 0)\)

- Both equations involve only squared coordinates. - Temporarily treat \(x^2\) and \(y^2\) as two unknowns in a linear system. - Recover all coordinate signs after finding the squared values.

1. Let \(X=x^2\) and \(Y=y^2\). The equations become \(\frac{X}{9}-\frac{Y}{16}=1\) and \(X+Y=25\). 2. Substitute \(Y=25-X\): \(\frac{X}{9}-\frac{25-X}{16}=1\). 3. Multiplying by \(144\) gives \(16X-9(25-X)=144\), so \(25X=369\) and \(X=\frac{369}{25}\). 4. Then \(Y=25-\frac{369}{25}=\frac{256}{25}\). 5. Therefore \(x=\pm\frac{3\sqrt{41}}{5}\) and \(y=\pm\frac{16}{5}\). Every independent sign combination satisfies both equations.

\((\pm\frac{3\sqrt{41}}{5}, \pm\frac{16}{5})\), with all four sign combinations

- Solve the hyperbola equation for both x-values at a fixed y-coordinate. - Subtract the left value from the right value. - Analyze the squared vertical displacement from the center to find minima and prescribed values.

1. Solve for the two x-values at a fixed height: \(x=1\pm3\sqrt{1+\frac{(y+2)^2}{16}}\). 2. Their separation is \(S(y)=6\sqrt{1+\frac{(y+2)^2}{16}}\). 3. This expression is minimized when \((y+2)^2=0\), so the minimum separation is \(6\) at \(y=-2\). 4. Set \(S(y)=8\): \(6\sqrt{1+\frac{(y+2)^2}{16}}=8\). 5. Squaring gives \(1+\frac{(y+2)^2}{16}=\frac{16}{9}\), so \((y+2)^2=\frac{112}{9}\). 6. Therefore \(y=-2\pm\frac{4\sqrt{7}}{3}\).

Minimum separation: \(6\), at \(y=-2\) Separation \(8\): \(y=-2+\frac{4\sqrt{7}}{3}\) and \(y=-2-\frac{4\sqrt{7}}{3}\)

- Compute the focal distance from the two denominator expressions. - Translate the asymptote angle into a slope condition. - Use that condition to determine the parameter value.

1. Here \(b^2=25-a^2\), so \(c^2=a^2+b^2=25\). Every member has foci \((\pm5, 0)\). 2. The asymptote slopes are \(\pm\frac{b}{a}=\pm\frac{\sqrt{25-a^2}}{a}\). 3. A \(45^\circ\) angle requires slope magnitude \(1\), so \(\sqrt{25-a^2}=a\). 4. Squaring gives \(25-a^2=a^2\), so \(a^2=\frac{25}{2}\). Then \(b^2=\frac{25}{2}\). 5. The required hyperbola is \(\frac{x^2}{25/2}-\frac{y^2}{25/2}=1\), with vertices \((\pm\frac{5}{\sqrt{2}}, 0)\).

Common foci: \((5, 0)\), \((-5, 0)\) Required equation: \(\frac{x^2}{25/2}-\frac{y^2}{25/2}=1\) Vertices: \((\frac{5}{\sqrt{2}}, 0)\), \((-\frac{5}{\sqrt{2}}, 0)\)

- Invert the coordinate transformation before substituting. - Read the transformed hyperbola’s features from standard form. - Apply the equal-denominator case carefully when finding asymptotes and foci.

1. Invert the transformation: \(x=\frac{X+1}{2}\) and \(y=\frac{Y-2}{3}\). 2. Substitution gives \(\frac{(X+1)^2}{36}-\frac{(Y-2)^2}{36}=1\). 3. The center is \((-1, 2)\), and \(a^2=b^2=36\). 4. The vertices are \((-1\pm6, 2)\), or \((-7, 2)\) and \((5, 2)\). 5. Since \(c^2=a^2+b^2=72\), the foci are \((-1\pm6\sqrt{2}, 2)\). 6. The asymptotes are \(Y-2=\pm(X+1)\).

Equation: \(\frac{(X+1)^2}{36}-\frac{(Y-2)^2}{36}=1\) Center: \((-1, 2)\) Vertices: \((-7, 2)\), \((5, 2)\) Foci: \((-1\pm6\sqrt{2}, 2)\) Asymptotes: \(Y-2=\pm(X+1)\)

- Write each asymptote in the form \(Ax+By+C=0\). - Apply the point-to-line distance formula to the same point on both asymptotes. - Multiply the two distances and use a scaled version of the hyperbola equation.

1. The asymptotes are \(y=\pm\frac{4}{3}x\), or \(4x-3y=0\) and \(4x+3y=0\). 2. The distances are \(d_1=\frac{|4x-3y|}{5}\) and \(d_2=\frac{|4x+3y|}{5}\). 3. Their product is \(d_1d_2=\frac{|16x^2-9y^2|}{25}\). 4. Multiplying the hyperbola equation by \(576\) gives \(16x^2-9y^2=576\). 5. Therefore, \(d_1d_2=\frac{576}{25}\), independent of \(P\).

\(d_1d_2=\frac{576}{25}\) square units for every point on the hyperbola.

- Substitute the line equation into the hyperbola and isolate the coefficient multiplying \(x^2\). - A real value of \(x\) requires the resulting value of \(x^2\) to be positive. - Compare the boundary slopes of the inequality with \(\pm\frac{b}{a}\).

1. Substitute \(y=mx\) into the hyperbola: \(\frac{x^2}{36}-\frac{m^2x^2}{16}=1\). 2. Factor \(x^2\): \(x^2\left(\frac{1}{36}-\frac{m^2}{16}\right)=1\). 3. Real intersection points require the coefficient of \(x^2\) to be positive: \(\frac{1}{36}-\frac{m^2}{16}>0\). 4. This inequality gives \(m^2<\frac{4}{9}\), so \(-\frac{2}{3}<m<\frac{2}{3}\). 5. The hyperbola's asymptotes are \(y=\pm\frac{2}{3}x\). At the boundary slopes, the substituted equation becomes \(0=1\), so the asymptotes do not meet the hyperbola. 6. Lines through the center with slopes strictly between the asymptote slopes meet both branches; steeper lines do not meet the hyperbola.

The line intersects the hyperbola exactly when \(-\frac{2}{3}<m<\frac{2}{3}\). The boundary slopes \(m=\pm\frac{2}{3}\) are the slopes of the asymptotes, which do not intersect the hyperbola.

- Group the x-terms and y-terms and complete the square in each variable. - Divide so the right side is \(1\), then identify the positive squared term to determine the opening direction. - Translate the asymptote equations \(y=\pm\frac{b}{a}x\) to the hyperbola's center.

1. Group the variable terms: \(4(x^2-4x)-9(y^2+6y)=101\). 2. Complete the squares: \(4((x-2)^2-4)-9((y+3)^2-9)=101\). 3. Simplify to obtain \(4(x-2)^2-9(y+3)^2=36\). 4. Divide by \(36\): \(\frac{(x-2)^2}{9}-\frac{(y+3)^2}{4}=1\). 5. The center is \((2,-3)\), and \(a=3\), so the vertices are \((-1,-3)\) and \((5,-3)\). 6. For a horizontal hyperbola, the asymptotes are \(y-k=\pm\frac{b}{a}(x-h)\). 7. Therefore, the asymptotes are \(y+3=\pm\frac{2}{3}(x-2)\).

Standard form: \(\frac{(x-2)^2}{9}-\frac{(y+3)^2}{4}=1\) Center: \((2, -3)\) Vertices: \((-1, -3)\) and \((5, -3)\) Asymptotes: \(y+3=\pm\frac{2}{3}(x-2)\)

- Use \(c^2=a^2+b^2\) to find the eccentricity and then the directrices. - Intersect each vertical directrix with both asymptotes. - Compute the squared distance of an intersection point from the center and use symmetry for the other three.

1. The hyperbola has \(a=8\), \(b=15\), and \(c=\sqrt{64+225}=17\), so its eccentricity is \(e=\frac{17}{8}\). 2. Its directrices are \(x=\pm\frac{a}{e}=\pm\frac{64}{17}\). 3. Its asymptotes are \(y=\pm\frac{15}{8}x\). 4. Intersecting the directrices with the asymptotes gives \(\left(\pm\frac{64}{17},\pm\frac{120}{17}\right)\), with the signs chosen independently. 5. Each point has squared distance from the origin \(\frac{4096}{289}+\frac{14400}{289}=64\). 6. Therefore, all four points lie on the circle \(x^2+y^2=64\).

Intersection points: \(\left(\pm\frac{64}{17}, \pm\frac{120}{17}\right)\), with signs chosen independently Common circle: \(x^2+y^2=64\)

- Write both the upper branch and the upper asymptote as formulas for \(y\). - At each specified x-coordinate, evaluate both formulas before subtracting their y-values. - Compare the two positive gaps to describe what “approaches an asymptote” means numerically.

1. The upper asymptote is \(y+2=\frac{5}{4}(x-1)\). 2. On the upper branch, \(y+2=5\sqrt{\frac{(x-1)^2}{16}-1}\). 3. At \(x=9\), the branch point is \((9,-2+5\sqrt{3})\), while the asymptote point is \((9,8)\). 4. Their vertical separation is \(8-(-2+5\sqrt{3})=10-5\sqrt{3}\). 5. At \(x=17\), the branch point is \((17,-2+5\sqrt{15})\), while the asymptote point is \((17,18)\). 6. Their vertical separation is \(18-(-2+5\sqrt{15})=20-5\sqrt{15}\). 7. Since \(20-5\sqrt{15}<10-5\sqrt{3}\), the branch is closer to the asymptote farther from the center, while never meeting it.

a) Branch point: \((9, -2+5\sqrt{3})\); asymptote point: \((9, 8)\); separation: \(10-5\sqrt{3}\) b) Branch point: \((17, -2+5\sqrt{15})\); asymptote point: \((17, 18)\); separation: \(20-5\sqrt{15}\) c) The separation decreases as \(x\) increases, illustrating that the branch approaches the asymptote.

- Use the asymptote slopes to determine the ratio \(\frac{b}{a}\). - Represent both semi-axis lengths with one positive scale factor, then substitute the known point. - The vertices of a horizontal hyperbola lie \(a\) units from the center on the x-axis.

1. A horizontal hyperbola centered at the origin has equation \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), with asymptote slopes \(\pm\frac{b}{a}\). 2. Since \(\frac{b}{a}=\frac{3}{2}\), write \(a=2k\) and \(b=3k\), where \(k>0\). 3. Substitute the given point: \(\frac{25}{4k^2}-\frac{81/4}{9k^2}=1\). 4. Simplifying gives \(\frac{4}{k^2}=1\), so \(k=2\). 5. Therefore, \(a=4\) and \(b=6\), and the equation is \(\frac{x^2}{16}-\frac{y^2}{36}=1\). 6. The vertices are \((-4,0)\) and \((4,0)\).

Equation: \(\frac{x^2}{16}-\frac{y^2}{36}=1\) Vertices: \((-4, 0)\) and \((4, 0)\)

- Apply the point-to-line distance formula to both asymptotes. - Split the absolute-value equation into its two possible linear cases. - Substitute each candidate line into the hyperbola and reject any case with no real points.

1. The distances are \(d_1=\frac{|4x-3y|}{5}\) and \(d_2=\frac{|4x+3y|}{5}\). 2. The condition \(d_1=2d_2\) becomes \(|4x-3y|=2|4x+3y|\). 3. One case is \(4x-3y=2(4x+3y)\), which simplifies to \(4x+9y=0\), or \(x=-\frac{9}{4}y\). 4. Substitution into the hyperbola gives \(\frac{y^2}{8}=1\), so \(y=\pm2\sqrt{2}\) and \(x=\mp\frac{9\sqrt{2}}{2}\). 5. The other case is \(4x-3y=-2(4x+3y)\), which gives \(y=-4x\). Substitution makes the left side of the hyperbola equation negative, so it gives no real points.

The points are \(\left(-\frac{9\sqrt{2}}{2}, 2\sqrt{2}\right)\) and \(\left(\frac{9\sqrt{2}}{2}, -2\sqrt{2}\right)\).

- Find the focal distance from the center to either focus. - For a vertical hyperbola, use \(\frac{a}{b}\) for the asymptote-slope magnitude. - Combine the ratio with \(c^2=a^2+b^2\) to determine both semi-axis parameters.

1. The focal distance is \(c=10\). 2. For a vertical hyperbola, the asymptote slopes are \(\pm\frac{a}{b}\). Thus \(\frac{a}{b}=\frac34\). 3. Write \(a=3k\) and \(b=4k\). Then \(c^2=a^2+b^2=9k^2+16k^2=25k^2\). 4. Since \(c=10\), \(5k=10\), so \(k=2\). Therefore \(a=6\) and \(b=8\). 5. The standard equation is \(\frac{(y-3)^2}{36}-\frac{(x+2)^2}{64}=1\). 6. The vertices are \((-2, 3\pm6)\), or \((-2, 9)\) and \((-2, -3)\). 7. The asymptotes are \(y-3=\pm\frac34(x+2)\).

Equation: \(\frac{(y-3)^2}{36}-\frac{(x+2)^2}{64}=1\) Vertices: \((-2, 9)\) and \((-2, -3)\) Asymptotes: \(y-3=\pm\frac34(x+2)\)

- Relate the asymptote slopes to the two semi-axis parameters. - Use the latus-rectum information as a second independent condition. - After finding the semi-axes, use the focal relationship for a hyperbola.

1. Write the hyperbola as \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), where \(a>0\) and \(b>0\). 2. The asymptote slope gives \(\frac{b}{a}=\frac{5}{12}\), so \(b=\frac{5a}{12}\). 3. A latus rectum has length \(\frac{2b^2}{a}\). Therefore, \(\frac{2}{a}\left(\frac{5a}{12}\right)^2=\frac{25}{6}\), which simplifies to \(a=12\). 4. Then \(b=5\), so the standard equation is \(\frac{x^2}{144}-\frac{y^2}{25}=1\). 5. Since \(c^2=a^2+b^2=169\), \(c=13\). The vertices are \((\pm12, 0)\), and the foci are \((\pm13, 0)\). 6. The eccentricity is \(e=\frac{13}{12}\), so the directrices are \(x=\pm\frac{a}{e}=\pm\frac{144}{13}\).

Standard equation: \(\frac{x^2}{144}-\frac{y^2}{25}=1\) Vertices: \((\pm12, 0)\) Foci: \((\pm13, 0)\) Directrices: \(x=\pm\frac{144}{13}\)

- Solve the hyperbola equation for the positive y-value and compare it with the positive-slope asymptote. - Rationalize the difference \(x-\sqrt{x^2-a^2}\) before solving the inequality. - Find the boundary where the gap equals \(0.2\), then use how the gap changes along the branch.

1. The upper branch is \(y=\frac{2}{3}\sqrt{x^2-36}\), and the upper-right asymptote is \(y=\frac{2}{3}x\). 2. At \(x=15\), the gap is \(10-\frac{2}{3}\sqrt{189}=10-2\sqrt{21}\approx0.835\). 3. For \(x\ge6\), the gap is \(D=\frac{2}{3}(x-\sqrt{x^2-36})\). 4. Rationalizing gives \(D=\frac{24}{x+\sqrt{x^2-36}}\). 5. Set the boundary value equal to \(0.2\): \(x+\sqrt{x^2-36}=120\). 6. Squaring \(\sqrt{x^2-36}=120-x\) gives \(x=\frac{1203}{20}\). The gap decreases as \(x\) increases, so it is below \(0.2\) when \(x>\frac{1203}{20}\).

a) \(10-2\sqrt{21}\approx0.835\) b) The gap is less than \(0.2\) for \(x>\frac{1203}{20}\approx60.15\).

- Translate the focal separation into a relationship involving the two semi-axis parameters. - Translate the guide rectangle's perimeter into a second relationship. - The given information may determine the two parameter values without determining which one is the transverse semi-axis.

1. The focal separation is \(2c=34\), so \(c=17\) and \(a^2+b^2=c^2=289\). 2. The guide rectangle has side lengths \(2a\) and \(2b\). Its perimeter is \(4(a+b)=92\), so \(a+b=23\). 3. Square the sum: \(529=a^2+2ab+b^2=289+2ab\), so \(ab=120\). 4. The positive numbers \(a\) and \(b\) are therefore the roots of \(z^2-23z+120=0\), giving \(\{a,b\}=\{8,15\}\). 5. If \(a=8\) and \(b=15\), the equation is \(\frac{x^2}{64}-\frac{y^2}{225}=1\), with vertices \((\pm8, 0)\) and asymptotes \(y=\pm\frac{15}{8}x\). 6. If \(a=15\) and \(b=8\), the equation is \(\frac{x^2}{225}-\frac{y^2}{64}=1\), with vertices \((\pm15, 0)\) and asymptotes \(y=\pm\frac{8}{15}x\).

Possibility 1: \(\frac{x^2}{64}-\frac{y^2}{225}=1\) Vertices: \((\pm8, 0)\) Asymptotes: \(y=\pm\frac{15}{8}x\) Possibility 2: \(\frac{x^2}{225}-\frac{y^2}{64}=1\) Vertices: \((\pm15, 0)\) Asymptotes: \(y=\pm\frac{8}{15}x\)