- 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 \(y-1=-\frac{5}{3}(x+2)\)

- The asymptotes of a hyperbola meet at its center. - The branch-opening direction tells you the transverse-axis orientation. - The vertices are the branch points closest to the center.

1. The two dashed asymptotes cross at \((-1, 1)\), so that point is the center. 2. The branches open left and right, so the transverse axis is horizontal. 3. The nearest points of the branches to the center lie at \((-3, 1)\) and \((1, 1)\). 4. Therefore those two points are the vertices.

Center: \((-1, 1)\) Transverse axis: horizontal Vertices: \((-3, 1)\) and \((1, 1)\)

- Use the point \(V\) to find the fixed absolute difference of focal distances. - For each other point, find its distance to each focus separately. - Why does exchanging the nearer and farther focus not change an absolute difference?

1. At \(V\), the distances to the foci are \(8\) and \(2\), so the required absolute difference is \(6\). 2. At \(P\), the distances are \(2\) and \(8\). Their absolute difference is also \(6\), so \(P\) is on the other branch. 3. At \(Q\), both distances are \(\sqrt{41}\), so the difference is \(0\). Thus \(Q\) is not on the hyperbola.

\(P=(-3,0)\) is on the hyperbola, on the opposite branch from \(V\). \(Q=(0,4)\) is not.

- Find two easy grid points on each dashed line. - Slope is vertical change divided by horizontal change. - Look at which directions the solid branches extend away from the center.

1. The dashed lines pass through the origin and the guide-rectangle corners \((3,2)\) and \((3,-2)\). 2. Their slopes are \(\frac{2}{3}\) and \(-\frac{2}{3}\). 3. The solid curves pass through \((-3,0)\) and \((3,0)\) and extend left and right, so the hyperbola opens horizontally.

Slopes: \(\frac{2}{3}\) and \(-\frac{2}{3}\) Opening: left and right

- Rearrange the equation to isolate the squared x-term. - A square can come from a positive or negative number. - Separate opening direction from the signs of individual coordinates.

1. Rearranging gives \(\frac{x^2}{16}=1+\frac{y^2}{9}\), so \(x^2\ge16\). 2. Thus \(x\le-4\) or \(x\ge4\). 3. The positive coefficient tells us the transverse axis is horizontal, not that both branches are on the right.

Rowan is incorrect. The branches satisfy \(x\le-4\) and \(x\ge4\); one branch lies left of the center.

- For each equation, which squared variable has a positive coefficient? - Find the asymptote equations for the two hyperbolas separately before comparing them.

1. The positive squared term determines the transverse axis: \(H_1\) opens left/right, while \(H_2\) opens up/down. 2. Thus \(H_2\) has vertices on the y-axis. 3. Both have the same guide rectangle: half-width \(3\) and half-height \(2\). Its diagonals give the shared asymptotes \(y=\pm\frac{2}{3}x\).

\(H_2\) has vertices on the y-axis. Both hyperbolas have asymptotes \(y=\frac{2}{3}x\) and \(y=-\frac{2}{3}x\).

- 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 \(y=-\frac{2}{3}x\)

- Determine the focal distance from the two hyperbola denominators. - Locate the right vertex on the transverse axis. - Compare the two vertex-to-focus distances with the hyperbola's constant difference.

1. The hyperbola has \(a=3\) and \(b=4\), so \(c^2=a^2+b^2=25\) and \(c=5\). 2. The foci are \((-5,0)\) and \((5,0)\), separated by \(10\) units. 3. The right vertex is \((3,0)\). 4. Its distance to the right focus is \(5-3=2\); its distance to the left focus is \(3-(-5)=8\). 5. The difference \(8-2=6\) equals \(2a\), as expected for a hyperbola.

Foci: \((-5,0)\), \((5,0)\) Focal separation: \(10\) Distances from right vertex: \(8\) to left focus and \(2\) to right focus

- 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

- 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: \(120\)

- Identify \(a\) and \(b\) from the denominators before using the asymptote formula. - For a horizontal hyperbola, compare the vertical change from the guide rectangle with the horizontal change. - Check whether the proposed slope is steeper or flatter than that rectangle suggests.

1. For a horizontal hyperbola \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\), the asymptotes are \(y-k=\pm\frac{b}{a}(x-h)\). 2. Here \(a=4\), \(b=3\), and the center is \((-2, 1)\). 3. Therefore the asymptotes are \(y-1=\pm\frac{3}{4}(x+2)\). 4. Jordan reversed the ratio \(\frac{b}{a}\).

Jordan is not correct. Correct asymptotes: \(y-1=\pm\frac{3}{4}(x+2)\) Mistake: Jordan reversed \(\frac{b}{a}\) to \(\frac{a}{b}\).

- Each linear factor measures a signed offset from an asymptote. - When is the product of two numbers negative? - Choose an x-coordinate on the requested side to find a sample y-coordinate.

1. The center is where the asymptotes meet: \((1,-2)\). 2. The product is negative, so the two factors have opposite signs. The branches are upper-left and lower-right of the center. 3. On the right branch choose \(x=3\), so \(2(y+2)=-6\) and \(y=-5\). Thus \((3,-5)\) lies on that branch.

The branches are upper-left and lower-right relative to \((1,-2)\), not upper-right and lower-left. One right-branch point is \((3,-5)\).

- For a hyperbola, relate the center-to-focus distance to both semi-axis lengths. - Compare the eccentricity with \(1\). - Recall how eccentricity determines the directrix position and check the order along the x-axis.

1. Here \(a=3\), \(b=4\), so \(c=\sqrt{a^2+b^2}=5\). 2. Thus the eccentricity is \(e=\frac{c}{a}=\frac{5}{3}\). 3. The right directrix is \(x=\frac{a}{e}=\frac{9}{5}\). 4. Because \(\frac{9}{5}<3\), it is closer to the center than the right vertex \((3,0)\).

\(e=\frac{5}{3}\); right directrix \(x=\frac{9}{5}\). Yes, it is closer to the center than the right vertex.

- Read the asymptote slope from the semi-axis lengths. - Match its positive magnitude with a familiar tangent value. - Use symmetry of the two lines about the x-axis.

1. The asymptote slopes are \(\pm\frac{b}{a}=\pm\frac{2}{\sqrt{12}}=\pm\frac{1}{\sqrt{3}}\). 2. Each asymptote makes a \(30^\circ\) angle with the positive x-axis, one above and one below it. 3. The smaller angle between the lines is \(30^\circ+30^\circ=60^\circ\).

\(60^\circ\)

- Locate the focus from the two semi-axis lengths. - Use different distance methods for the vertex and the off-axis point. - Check the second point in the equation before comparing its two distances.

1. Since \(a=4\), \(b=3\), and \(c=\sqrt{a^2+b^2}=5\), the right focus is \((5,0)\), the right vertex is \(V=(4,0)\), and \(e=\frac{5}{4}\). 2. At \(V\), the focus distance is \(1\), while the directrix distance is \(4-\frac{16}{5}=\frac{4}{5}\). Their ratio is \(\frac{5}{4}\). 3. At \(P=(\frac{20}{3},4)\), \(\frac{(20/3)^2}{16}-\frac{4^2}{9}=\frac{25}{9}-\frac{16}{9}=1\), so \(P\) belongs to the hyperbola. 4. The focus distance is \(\sqrt{(\frac{20}{3}-5)^2+4^2}=\frac{13}{3}\). The directrix distance is \(\frac{20}{3}-\frac{16}{5}=\frac{52}{15}\). 5. Their ratio is \(\frac{13/3}{52/15}=\frac{5}{4}\), the same as at the vertex and equal to \(e\).

Right focus: \((5,0)\). Both \(V=(4,0)\) and \(P=(\frac{20}{3},4)\) give distance ratio \(\frac{5}{4}=e\); \(P\) satisfies the hyperbola equation.

- 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 \((4, 0)\)

- 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.

- 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: \(x=\pm\frac{144}{13}\)

- 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: \(2\arctan\left(\frac47\right)\approx59.5^\circ\)

- Solve for the positive y-value on the hyperbola and compare it with the positive-slope asymptote. - Rationalize the curve–line difference to simplify it. - Find the boundary where the gap equals \(0.2\), then consider how the gap changes along the branch.

1. The upper branch is \(y=\frac{2}{3}\sqrt{x^2-36}\), and the 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 \(D=\frac{2}{3}(x-\sqrt{x^2-36})=\frac{24}{x+\sqrt{x^2-36}}\). 4. At the boundary \(D=0.2\), so \(x+\sqrt{x^2-36}=120\). 5. Solving \(\sqrt{x^2-36}=120-x\) gives \(x=\frac{1203}{20}\). 6. The gap decreases as \(x\) increases, so it is below \(0.2\) for \(x>\frac{1203}{20}\).

a) \(10-2\sqrt{21}\approx0.835\) b) \(x>\frac{1203}{20}=60.15\)

- 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: \(y=\pm\frac{8}{15}x\)