Free math worksheets

- Compare which conics are bounded and which extend indefinitely. - Use the statement that the star lies at a focus to rule out a circle.

1. A bounded noncircular conic is an ellipse rather than a parabola or hyperbola. 2. An ellipse has two foci, and an orbital model may place the star at one of them. 3. Therefore an ellipse matches the stated orbital geometry.

Ellipse; it is a bounded conic with foci, and the star can lie at one focus.

- Distinguish a physical object from a mathematical model of that object. - Ask what the words “modeled by” authorize you to assume in the calculation.

1. A mathematical model is an intentional approximation or idealization of a physical object. 2. Saying that the cable is modeled by a parabola does not claim that every real cable is exactly parabolic. 3. It means that, for this problem, the cable's cross section is assumed to follow a parabola closely enough that parabola equations and properties may be used.

No. The statement is a modeling assumption: for this problem, the cable is approximated by a parabola, so parabola equations and properties may be used.

- Convert the time difference into a difference in travel distances. - Think about the largest possible difference between distances to two fixed points. - Use the sensor separation to determine the limiting delay.

1. A time difference of \(0.070\,\text{s}\) would require a difference in travel distances of \(340(0.070)=23.8\,\text{m}\). 2. For any point, the absolute difference of its distances from two fixed sensors cannot exceed the distance between the sensors. Here that maximum distance difference is \(20\,\text{m}\). 3. Because \(23.8>20\), no source location can produce the reported time difference. 4. The greatest possible time difference is \(\frac{20}{340}=\frac{1}{17}\,\text{s}\approx0.0588\,\text{s}\).

No. The reported delay would require a distance difference of \(23.8\,\text{m}\), which exceeds the \(20\,\text{m}\) separation of the sensors. The greatest possible arrival-time difference is \(\frac{1}{17}\,\text{s}\approx0.0588\,\text{s}\).

- Use half the cooker's width as the horizontal coordinate of a rim point. - Relate that point to the focal parameter in a vertical parabola. - Compare the calculated focus height with the current receiver height.

1. Model the cross section by \(x^2=4py\), where the focus is \((0,p)\). 2. The rim is \(0.90\,\text{m}\) from the axis and \(0.225\,\text{m}\) above the vertex, so \((0.90)^2=4p(0.225)\). 3. Solving gives \(p=0.90\,\text{m}\). 4. The receiver is at height \(0.80\,\text{m}\), which is \(0.10\,\text{m}\) below the focus. It must be raised by \(0.10\,\text{m}\).

The focal distance is \(0.90\,\text{m}\). The receiver must be moved upward by \(0.10\,\text{m}\).

- Convert the full gallery dimensions to semi-axis lengths. - Use the relationship among an ellipse's semi-axes and focal distance. - Recall what quantity stays constant for every point on an ellipse.

1. The horizontal semi-axis is \(a=15\,\text{ft}\), and the vertical semi-axis is \(b=9\,\text{ft}\). 2. The focal distance satisfies \(c^2=a^2-b^2=225-81=144\), so \(c=12\,\text{ft}\). 3. Since the ellipse is centered at the origin with a horizontal major axis, the listening points should be placed at the foci \((-12,0)\) and \((12,0)\). 4. For every point on an ellipse, the sum of its distances to the two foci is \(2a\). 5. Therefore the total length of the two straight segments is \(2a=30\,\text{ft}\).

Place the listening points at \((-12,0)\) and \((12,0)\). The total two-segment path length is always \(30\,\text{ft}\).

- Use the closest branch point to determine the transverse semi-axis. - Convert eccentricity into focal distance, then find the remaining semi-axis. - Interpret asymptote slopes as long-range direction angles.

1. The vertex gives \(a=5\). The eccentricity gives \(c=ea=13\). 2. For a hyperbola, \(b^2=c^2-a^2=169-25=144\), so \(b=12\). 3. The standard equation is \(\frac{x^2}{25}-\frac{y^2}{144}=1\). The specified trajectory is the right branch, so \(x\ge5\). 4. The asymptotes are \(y=\pm\frac{b}{a}x=\pm\frac{12}{5}x\). 5. Their direction angles are \(\theta=\pm\arctan(\frac{12}{5})\approx\pm67.38^\circ\).

Trajectory: \(\frac{x^2}{25}-\frac{y^2}{144}=1\), restricted to the right branch \(x\ge5\) Asymptotic direction angles: \(\pm\arctan(\frac{12}{5})\approx\pm67.38^\circ\)

- Read the full width and maximum height from the diagram and convert them to ellipse semi-axis lengths. - Model the roof cross section with the upper half of a standard ellipse centered at floor level. - At the stated beam height, solve for the two symmetric horizontal coordinates and use their separation.

1. From the diagram, the semiellipse has horizontal semi-axis \(a=30\,\text{ft}\) and vertical semi-axis \(b=24\,\text{ft}\). 2. Its upper half satisfies \(\frac{x^2}{900}+\frac{y^2}{576}=1\). 3. At beam height \(y=20\), \(\frac{x^2}{900}+\frac{400}{576}=1\). 4. Since \(\frac{400}{576}=\frac{25}{36}\), \(\frac{x^2}{900}=\frac{11}{36}\), so \(x=\pm5\sqrt{11}\). 5. The full horizontal span is twice the positive x-coordinate: \(10\sqrt{11}\,\text{ft}\approx33.2\,\text{ft}\).

\(10\sqrt{11}\,\text{ft}\approx33.2\,\text{ft}\)

- Interpret the stakes as the two foci. - Relate their separation and the fixed distance sum to \(c\) and \(a\). - Use the standard ellipse relationship to find the remaining semi-axis.

1. The stakes are the foci, so \(2c=20\) and \(c=10\). 2. The constant focal-distance sum is \(2a=26\), so \(a=13\). 3. Then \(b^2=a^2-c^2=169-100=69\), so \(b=\sqrt{69}\). 4. The major-axis length is \(2a=26\,\text{ft}\), and the minor-axis length is \(2b=2\sqrt{69}\,\text{ft}\). 5. The equation is \(\frac{x^2}{169}+\frac{y^2}{69}=1\).

Major-axis length: \(26\,\text{ft}\) Minor-axis length: \(2\sqrt{69}\,\text{ft}\) Equation: \(\frac{x^2}{169}+\frac{y^2}{69}=1\)

- Determine the focal distance from the two hyperbola denominators. - Locate the vertex on the branch used by the reflector. - 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-hand vertex is \((3, 0)\). 4. Its distance to the right focus is \(5-3=2\), and its distance to the left focus is \(3-(-5)=8\). 5. The difference \(8-2=6\) equals \(2a\), consistent with the hyperbola’s defining property.

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

- Separate the optical or acoustic reflection conditions from the distance-locus condition. - Recall which conic has one focus and a single axis-directed reflection property. - Compare the sum and difference versions of the two-focus definitions.

1. A parabola reflects rays parallel to its axis toward its focus, so design a) uses a parabola. 2. An ellipse reflects a ray leaving one focus toward the other focus, so design b) uses an ellipse. 3. A hyperbola is the locus of points for which the absolute difference of the distances to two foci is constant, so design c) uses a hyperbola.

a) Parabola — parallel incoming rays reflect toward the focus. b) Ellipse — a ray from one focus reflects toward the other focus. c) Hyperbola — the absolute difference of the distances to the two foci is constant.

- Identify which direction is unchanged by the tilt. - Determine the projection factor for lengths perpendicular to the tilt axis. - Use the two projected semi-axis lengths to write the ellipse and find its area.

1. Orthogonal projection preserves lengths parallel to the tilt axis, so the projected semi-axis along the x-axis is \(6\,\text{ft}\). 2. Lengths perpendicular to the tilt axis are multiplied by \(\cos 60^\circ=\frac{1}{2}\), so the other projected semi-axis is \(6\cdot\frac{1}{2}=3\,\text{ft}\). 3. The projected boundary is the ellipse \(\frac{x^2}{36}+\frac{y^2}{9}=1\). 4. Its area is \(\pi ab=\pi(6)(3)=18\pi\,\text{ft}^2\).

Projected boundary: \(\frac{x^2}{36}+\frac{y^2}{9}=1\) Projected area: \(18\pi\,\text{ft}^2\)

- Compare each plane-axis angle with the generator-axis angle. - Equality corresponds to a plane parallel to a generator. - Decide whether the plane cuts one nappe or both nappes.

1. A plane perpendicular to the axis produces a circle, so cut a) is a circle. 2. When the plane-axis angle is greater than the generator-axis angle but less than \(90^\circ\), the plane cuts one nappe in an ellipse. Thus, cut b) is an ellipse. 3. When the plane is parallel to a generator, the angles are equal and the section is a parabola. Thus, cut c) is a parabola. 4. When the plane-axis angle is less than the generator-axis angle, the plane cuts both nappes and produces a hyperbola. Thus, cut d) is a hyperbola.

a) circle b) ellipse c) parabola d) hyperbola

- Start with a parabola written in vertex form. - Use the nozzle location to determine the remaining coefficient. - Set the height equal to zero and select the physically relevant solution.

1. Use vertex form \(y=a(x-3)^2+5\). 2. Substitute the nozzle point \((0, 1)\): \(1=9a+5\), so \(a=-\frac{4}{9}\). 3. The path is \(y=-\frac{4}{9}(x-3)^2+5\). 4. Set \(y=0\) to find the pool-level intersections: \((x-3)^2=\frac{45}{4}\). 5. Thus, \(x=3\pm\frac{3\sqrt{5}}{2}\). The positive landing coordinate is \(3+\frac{3\sqrt{5}}{2}\).

Path: \(y=-\frac{4}{9}(x-3)^2+5\) Landing point: \(\left(3+\frac{3\sqrt{5}}{2}, 0\right)\), approximately \((6.35, 0)\)

- Use the wheel's center and radius in the standard equation of a circle. - Substitute the specified height for \(y\) before solving for the horizontal coordinate. - The two square roots represent positions on opposite sides of the vertical centerline.

1. The circle has center \((0,20)\) and radius \(18\), so its equation is \(x^2+(y-20)^2=324\). 2. At a height of \(29\,\text{ft}\), substitute \(y=29\): \(x^2+9^2=324\). 3. Thus, \(x^2=243\), so \(x=\pm9\sqrt{3}\). 4. Therefore, the rider is about \(15.6\,\text{ft}\) to either side of the centerline.

a) \(x^2+(y-20)^2=324\) b) \(x=\pm9\sqrt{3}\,\text{ft}\), approximately \(15.6\,\text{ft}\) to either side of the centerline

- Use the focal length to write the parabola's cross-sectional equation. - Read the aperture radius from the diagram and evaluate the parabola there. - Substitute the radius and depth into the supplied volume formula.

1. With focal length \(p=2\), the cross section is \(x^2=4py=8y\). 2. From the diagram, the rim is at \(x=3\,\text{m}\), so the depth is \(d=y=\frac{9}{8}\,\text{m}\). 3. The receiver belongs at the focus, \((0,2)\), which is \(2\,\text{m}\) above the vertex. 4. The volume is \(V=\frac{1}{2}\pi(3)^2\left(\frac{9}{8}\right)=\frac{81\pi}{16}\,\text{m}^3\).

Rim depth: \(\frac{9}{8}\,\text{m}\) Receiver location: \((0,2)\) Volume: \(\frac{81\pi}{16}\,\text{m}^3\)

- One axis of the oblique cut remains the pipe's diameter. - Relate the elongated axis to its perpendicular projection. - Use the resulting semi-axis lengths in the ellipse area formula.

1. The minor axis is perpendicular to the direction of elongation and remains equal to the pipe diameter, so its length is \(8\,\text{in.}\). 2. Let \(L\) be the major-axis length. Its projection onto a plane perpendicular to the pipe axis equals the diameter: \(L\sin30^\circ=8\). 3. Therefore, \(L=16\,\text{in.}\). 4. The semi-axis lengths are \(a=8\,\text{in.}\) and \(b=4\,\text{in.}\). 5. The opening area is \(\pi ab=32\pi\,\text{in.}^2\).

Major axis: \(16\,\text{in.}\) Minor axis: \(8\,\text{in.}\) Area: \(32\pi\,\text{in.}^2\)

- Apply the supplied approximation separately to the two pairs of semi-axes. - Keep the two perimeter calculations organized before subtracting. - Round only after evaluating each full expression.

1. For the inner lane, \(C_{\text{in}}\approx\pi\left[3(60+40)-\sqrt{(180+40)(60+120)}\right]\). 2. This simplifies to \(C_{\text{in}}\approx60\pi(5-\sqrt{11})\approx317.31\,\text{m}\). 3. For the outer lane, \(C_{\text{out}}\approx\pi\left[3(64+44)-\sqrt{(192+44)(64+132)}\right]\). 4. This simplifies to \(C_{\text{out}}\approx4\pi(81-7\sqrt{59})\approx342.21\,\text{m}\). 5. The difference is approximately \(342.21-317.31=24.90\,\text{m}\).

Inner-lane lap: approximately \(317.31\,\text{m}\) Outer-lane lap: approximately \(342.21\,\text{m}\) Outer-lane increase: approximately \(24.90\,\text{m}\)

- Use the diagram to place the vertex and the two floor-level endpoints in a convenient coordinate system. - Determine the coefficient in a vertex-form parabola from one endpoint. - For the widest centered doorway of the stated height, its two upper corners lie on the arch.

1. Put the arch endpoints at \((-12,0)\) and \((12,0)\), with vertex \((0,9)\). 2. In vertex form, \(y=9+ax^2\). Using \((12,0)\) gives \(0=9+144a\), so \(a=-\frac{1}{16}\). 3. Therefore the arch is \(y=9-\frac{x^2}{16}\). 4. The upper corners of the tallest possible \(6\,\text{ft}\)-high centered doorway lie on the arch, so set \(y=6\): \(6=9-\frac{x^2}{16}\). 5. Then \(x^2=48\), so the corners have \(x=\pm4\sqrt{3}\). 6. The maximum doorway width is the horizontal separation \(8\sqrt{3}\,\text{ft}\approx13.9\,\text{ft}\).

\(8\sqrt{3}\,\text{ft}\approx13.9\,\text{ft}\)

- Locate both foci from the semi-axis lengths. - Use the constant sum of focal distances before substituting coordinates. - What line contains points equidistant from the two foci?

1. The focal distance is \(c=\sqrt{100-64}=6\), so the star is at \((6, 0)\). 2. The sum of the distances from an ellipse point to the two foci is \(2a=20\). 3. If the distance to the right focus is \(10\), then the distance to the left focus is also \(10\). Therefore the point lies on the perpendicular bisector of the focal segment, so \(x=0\). 4. Substituting \(x=0\) into the ellipse gives \(\frac{y^2}{64}=1\), so \(y=\pm8\).

The probe can be at \((0, 8)\) or \((0, -8)\), in millions of miles.

- Halve the focal separation and major-axis length to obtain \(c\) and \(a\). - Use the ellipse relation \(a^2=b^2+c^2\) to find the other semi-axis. - The defining focal-distance sum of an ellipse equals its full major-axis length.

1. Half the focal separation is \(c=15\,\text{cm}\), and half the major-axis length is \(a=25\,\text{cm}\). 2. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=625-225=400\) and \(b=20\,\text{cm}\). 3. The standard equation is \(\frac{x^2}{625}+\frac{y^2}{400}=1\). 4. The minor-axis length is \(2b=40\,\text{cm}\). 5. The sum of the focal distances is \(2a=50\,\text{cm}\).

a) \(\frac{x^2}{625}+\frac{y^2}{400}=1\) b) \(40\,\text{cm}\) c) \(50\,\text{cm}\)

- Use the positive y-value because only the upper half models the roof. - For a horizontal cross section, substitute the given height and solve for both x-coordinates. - The total width is the distance between the left and right intersection coordinates.

1. Substitute \(x=8\): \(\frac{64}{100}+\frac{y^2}{36}=1\). 2. Since the roof is the upper half, \(y=\frac{18}{5}=3.6\). This is the support height. 3. For the width at \(y=3\), substitute \(y=3\): \(\frac{x^2}{100}+\frac{9}{36}=1\). 4. Thus, \(\frac{x^2}{100}=\frac{3}{4}\), so \(x=\pm5\sqrt{3}\). 5. The total width is \(10\sqrt{3}\,\text{ft}\approx17.3\,\text{ft}\).

a) \(\frac{18}{5}\,\text{ft}=3.6\,\text{ft}\) b) \(10\sqrt{3}\,\text{ft}\approx17.3\,\text{ft}\)

- Read the two projected diameters from the diagram. - One diameter is unchanged because it lies along the tilt axis; relate the shortened diameter to it with a cosine factor. - Use the two projected semi-axis lengths for the ellipse area.

1. Under orthographic projection, the diameter parallel to the tilt axis keeps its full length. From the diagram, the original circle's diameter is \(12\,\text{cm}\), so its radius is \(6\,\text{cm}\). 2. If \(\alpha\) is the angle between the two planes, the perpendicular diameter is shortened by the factor \(\cos\alpha\). 3. The diagram gives \(12\cos\alpha=7.2\), so \(\cos\alpha=0.6\). 4. Therefore \(\alpha=\arccos(0.6)\approx53.1^\circ\). 5. The projected ellipse has semi-axes \(6\,\text{cm}\) and \(3.6\,\text{cm}\), so its area is \(\pi(6)(3.6)=21.6\pi\,\text{cm}^2\).

Original radius: \(6\,\text{cm}\) Tilt angle: \(\arccos(0.6)\approx53.1^\circ\) Projected area: \(21.6\pi\,\text{cm}^2\)

- Read the full ground-level span and the center height from the diagram, then convert the span to a horizontal semi-axis while using the center height as the vertical semi-axis. - Substitute the given horizontal position and select the nonnegative y-value for the arch. - At the specified height, find both symmetric x-coordinates before computing the width.

1. From the diagram, the horizontal semi-axis is \(a=15\,\text{ft}\), and the vertical semi-axis is \(b=5\,\text{ft}\). 2. The complete ellipse is \(\frac{x^2}{225}+\frac{y^2}{25}=1\). The arch is its upper half, so \(y\ge0\). 3. At \(x=9\), \(\frac{81}{225}+\frac{y^2}{25}=1\). 4. Therefore \(\frac{y^2}{25}=\frac{16}{25}\), so \(y=4\) on the upper half. 5. At height \(y=3\), \(\frac{x^2}{225}+\frac{9}{25}=1\), so \(\frac{x^2}{225}=\frac{16}{25}\) and \(x=\pm12\). 6. The horizontal clearance width is \(24\,\text{ft}\).

a) \(\frac{x^2}{225}+\frac{y^2}{25}=1\), with \(y\ge0\) for the arch b) \(4\,\text{ft}\) c) \(24\,\text{ft}\)

- Compare the words “sum” and “absolute difference” with the focal definitions of conics. - Treat the two systems separately before comparing them. - The conditions on the constants rule out degenerate cases.

1. An ellipse is the locus of points for which the sum of the distances to two fixed foci is constant. Therefore System A produces an ellipse. 2. A hyperbola is the locus of points for which the absolute difference of the distances to two fixed foci is constant. Therefore System B produces a hyperbola. 3. The stated size conditions on the constants make the two loci nondegenerate.

a) Ellipse; the sum of the distances to the two foci is constant. b) Hyperbola; the absolute difference of the distances to the two foci is constant.

- Interpret the fixed difference between distances to two fixed points as a conic condition. - Compare half the focal separation with half the stated distance difference. - Use which tower is farther away to select the correct branch before substituting the x-coordinate.

1. The towers are the foci, so \(c=6\). The constant difference of focal distances is \(2a=8\), so \(a=4\). 2. For a horizontal hyperbola, \(c^2=a^2+b^2\). Thus \(b^2=36-16=20\). 3. Because the strike is farther from the western tower, it lies on the eastern branch. The branch is part of \(\frac{x^2}{16}-\frac{y^2}{20}=1\) with \(x>0\). 4. Substituting \(x=10\) gives \(\frac{100}{16}-\frac{y^2}{20}=1\), so \(y^2=105\) and \(y=\pm\sqrt{105}\). 5. The two locations are \((10, \sqrt{105})\) and \((10, -\sqrt{105})\), approximately \((10, 10.25)\) and \((10, -10.25)\).

a) \(\frac{x^2}{16}-\frac{y^2}{20}=1\), restricted to the branch \(x>0\). b) \((10, \sqrt{105})\) and \((10, -\sqrt{105})\), approximately \((10, 10.25)\) and \((10, -10.25)\), in miles.

- Relate the two extreme distances from one focus to the semi-major axis and focal distance. - Add and subtract those two relationships. - Use the ellipse relationship among the two semi-axes and focal distance.

1. For a horizontal ellipse measured from the right focus, the nearest and farthest focal distances are \(a-c\) and \(a+c\). 2. Thus \(a-c=7000\) and \(a+c=13{,}000\). Adding gives \(a=10{,}000\), and subtracting gives \(c=3000\). 3. Then \(b^2=a^2-c^2=100{,}000{,}000-9{,}000{,}000=91{,}000{,}000\). 4. The equation is \(\frac{x^2}{100000000}+\frac{y^2}{91000000}=1\). 5. The foci are \((-3000, 0)\) and \((3000, 0)\), and the eccentricity is \(e=\frac{c}{a}=0.3\).

\(\frac{x^2}{100000000}+\frac{y^2}{91000000}=1\) Foci: \((-3000, 0)\) and \((3000, 0)\) Eccentricity: \(0.3\)

- Use vertex form with the cable's lowest point as the vertex. - Use one tower point to determine the remaining coefficient. - Evaluate the model at the listed symmetric positions before adding hanger lengths.

1. With the vertex at \((0,20)\), write the parabolic model as \(y=ax^2+20\). 2. A tower point is \((200,100)\), so \(100=40{,}000a+20\), giving \(a=\frac{1}{500}\). 3. The cable equation is \(y=\frac{x^2}{500}+20\). 4. The hanger lengths are \(20\) at \(x=0\), \(25\) at \(x=\pm50\), \(40\) at \(x=\pm100\), and \(65\) at \(x=\pm150\). 5. Their total length is \(20+2(25+40+65)=280\,\text{ft}\).

Cable model: \(y=\frac{x^2}{500}+20\) Total hanger length: \(280\,\text{ft}\)

- Read the two full widths from the diagram and convert them to half-widths. - The narrowest half-width determines the transverse semi-axis. - Use the point at height \(40\,\text{ft}\) to determine the remaining scale, then evaluate at the requested height.

1. Use \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), where \(y\) measures height from the narrowest level. 2. The minimum half-width is \(a=30\,\text{ft}\), so \(a^2=900\). 3. The point \((50,40)\) lies on the profile. Substitution gives \(\frac{2500}{900}-\frac{1600}{b^2}=1\), so \(b^2=900\). 4. The model is \(\frac{x^2}{900}-\frac{y^2}{900}=1\). 5. At \(y=60\), \(\frac{x^2}{900}-4=1\), so \(x=30\sqrt{5}\) on the right side. 6. The full width is \(2x=60\sqrt{5}\,\text{ft}\approx134.16\,\text{ft}\).

Model: \(\frac{x^2}{900}-\frac{y^2}{900}=1\) Predicted width \(60\,\text{ft}\) above the narrowest level: \(60\sqrt{5}\,\text{ft}\approx134.16\,\text{ft}\)

- Write the upper half of the ellipse using half the tunnel width and the center height. - Because the vehicle is centered, determine the horizontal coordinate of either top corner. - Find the tunnel height at that horizontal coordinate and compare it with the vehicle height.

1. The tunnel has horizontal semi-axis \(a=20\,\text{ft}\) and vertical semi-axis \(b=12\,\text{ft}\), so its upper boundary satisfies \(\frac{x^2}{400}+\frac{y^2}{144}=1\), with \(y\ge0\). 2. Because the \(14\,\text{ft}\)-wide vehicle is centered, its top corners have \(x=\pm7\). Their height is \(11\,\text{ft}\). 3. At \(x=7\), the tunnel height is \(y=12\sqrt{1-\frac{49}{400}}=\frac{3\sqrt{351}}{5}\,\text{ft}\). 4. Since \(\frac{3\sqrt{351}}{5}\approx11.241>11\), the vehicle clears the tunnel. 5. The minimum clearance occurs above the two top corners: \(\frac{3\sqrt{351}}{5}-11\approx0.241\,\text{ft}\).

Yes. The minimum vertical clearance is \(\frac{3\sqrt{351}}{5}-11\,\text{ft}\approx0.241\,\text{ft}\).

- Fit each model from the shared center and endpoint data. - Evaluate both equations at the same horizontal location. - Compare the two resulting heights rather than only comparing their equations.

1. Model A has form \(y=8-ax^2\). Using \((10, 0)\) gives \(a=\frac{2}{25}\), so \(y=8-\frac{2}{25}x^2\). 2. Model B has semi-axes \(a=10\) and \(b=8\), so \(\frac{x^2}{100}+\frac{y^2}{64}=1\), with \(y\ge0\). 3. At \(x=6\), Model A has height \(8-\frac{72}{25}=\frac{128}{25}=5.12\). 4. Model B has height \(8\sqrt{1-\frac{36}{100}}=\frac{32}{5}=6.4\). 5. The semielliptical model is higher by \(\frac{32}{5}-\frac{128}{25}=\frac{32}{25}=1.28\).

Model A: \(y=8-\frac{2}{25}x^2\) Model B: \(\frac{x^2}{100}+\frac{y^2}{64}=1\), \(y\ge0\) At \(x=6\), Model B gives \(1.28\) more units of clearance.

- Identify the focus and directrix from the parabola equation. - Express the incoming horizontal distance in terms of the reflection point’s x-coordinate. - Replace the focus distance using the defining property of a parabola.

1. The parabola has \(4p=16\), so its focus is \(F=(4, 0)\) and its directrix is \(x=-4\). 2. The incoming horizontal segment from the wavefront point \((12, y)\) to \(P=(x, y)\) has length \(12-x\). 3. By the focus–directrix definition of the parabola, \(PF\) equals the perpendicular distance from \(P\) to \(x=-4\), which is \(x+4\). 4. The total distance is \((12-x)+(x+4)=16\,\text{m}\), independent of \(P\). 5. The travel time is \(\frac{16}{343}\,\text{s}\approx0.0466\,\text{s}\), or approximately \(46.6\,\text{ms}\).

Every reflected path has total length \(16\,\text{m}\). Travel time: \(\frac{16}{343}\,\text{s}\approx0.0466\,\text{s}\), or approximately \(46.6\,\text{ms}\)

- Represent a boundary point on the follower ray using polar-style coordinates centered at the ellipse's center. - Substitute the ray coordinates into the ellipse equation. - After the cam rotates, update the angle between the fixed follower direction and the major axis.

1. Along a ray making angle \(\theta\) with the major axis, write \((x, y)=(r\cos\theta,r\sin\theta)\). 2. Substitution gives \(r=\frac{1}{\sqrt{\frac{\cos^2\theta}{100}+\frac{\sin^2\theta}{36}}}\). 3. At \(\theta=30^\circ\), \(r=\frac{1}{\sqrt{\frac{3}{400}+\frac{1}{144}}}=\frac{30}{\sqrt{13}}\approx8.32\). 4. After a \(90^\circ\) rotation, the fixed follower direction makes an angle of \(60^\circ\) in magnitude with the major axis. 5. Thus, \(r=\frac{1}{\sqrt{\frac{1}{400}+\frac{3}{144}}}=\sqrt{\frac{300}{7}}=\frac{10\sqrt{21}}{7}\approx6.55\). 6. The follower moves inward by \(\frac{30}{\sqrt{13}}-\frac{10\sqrt{21}}{7}\approx1.77\) units.

Initial distance: \(\frac{30}{\sqrt{13}}\approx8.32\) units Distance after rotation: \(\frac{10\sqrt{21}}{7}\approx6.55\) units Inward change: approximately \(1.77\) units

- Write the translated ellipse equation from the center and two semi-axis lengths. - Substitute the line equation into the ellipse and solve the resulting quadratic. - Use the distance formula between the two boundary crossings, keeping the exact fractional coordinates until the end.

1. The boundary of the no-fly zone is \(\frac{(x-4)^2}{100}+\frac{(y+2)^2}{36}=1\). 2. Substitute the flight path \(y=x\): \(\frac{(x-4)^2}{100}+\frac{(x+2)^2}{36}=1\). 3. Multiplying by \(900\) and simplifying gives \(17x^2+14x-328=0\). 4. The roots are \(x=4\) and \(x=-\frac{82}{17}\). Since \(y=x\), the boundary points are \((4,4)\) and \(\left(-\frac{82}{17},-\frac{82}{17}\right)\). 5. The change in each coordinate is \(4+\frac{82}{17}=\frac{150}{17}\). 6. The distance traveled inside the ellipse is \(\sqrt{\left(\frac{150}{17}\right)^2+\left(\frac{150}{17}\right)^2}=\frac{150\sqrt{2}}{17}\,\text{km}\approx12.5\,\text{km}\).

Entry and exit points: \(\left(-\frac{82}{17}, -\frac{82}{17}\right)\) and \((4, 4)\) Distance inside the zone: \(\frac{150\sqrt{2}}{17}\,\text{km}\approx12.5\,\text{km}\)

- Interpret the two tracking stations as the foci of the locus. - Relate the constant difference of the focal distances to the transverse semi-axis. - Use the relationship among \(a\), \(b\), and \(c\) for a hyperbola. - After finding the locus, use the given y-coordinate and the fact that the vessel is on the right-hand branch.

1. The stations are the foci \((-5,0)\) and \((5,0)\), so \(c=5\). 2. For a hyperbola, the constant difference of the focal distances is \(2a\). Thus \(2a=6\), so \(a=3\). 3. Since \(c^2=a^2+b^2\), \(b^2=25-9=16\). 4. Therefore the hyperbolic locus has equation \(\frac{x^2}{9}-\frac{y^2}{16}=1\). 5. Substitute \(y=12\): \(\frac{x^2}{9}-\frac{144}{16}=1\). 6. Hence \(\frac{x^2}{9}-9=1\), so \(x^2=90\). 7. Therefore \(x=\pm3\sqrt{10}\). Since the vessel is on the right-hand branch, \(x=3\sqrt{10}\). 8. The vessel is at \((3\sqrt{10},12)\). 9. Its distance from \(A\) is \(\sqrt{(3\sqrt{10}+5)^2+12^2}=5\sqrt{10}+3\). 10. Its distance from \(B\) is \(\sqrt{(3\sqrt{10}-5)^2+12^2}=5\sqrt{10}-3\). 11. The difference is \((5\sqrt{10}+3)-(5\sqrt{10}-3)=6\).

a) \(\frac{x^2}{9}-\frac{y^2}{16}=1\) b) \((3\sqrt{10},12)\) c) The distances are \(5\sqrt{10}+3\,\text{km}\) and \(5\sqrt{10}-3\,\text{km}\), whose difference is \(6\,\text{km}\).

- Use the vertex and focus distances to determine \(a\), \(c\), and then \(b\). - Substitute the given x-coordinate and choose the positive y-value for the upper point. - Compute both focal distances and compare their difference with \(2a\).

1. The center-to-vertex distance is \(a=5\), and the focal distance is \(c=13\). 2. For a hyperbola, \(c^2=a^2+b^2\), so \(b^2=169-25=144\). 3. The standard equation is \(\frac{x^2}{25}-\frac{y^2}{144}=1\), and the asymptotes are \(y=\pm\frac{12}{5}x\). 4. At \(x=13\), \(\frac{169}{25}-\frac{y^2}{144}=1\). 5. Thus \(\frac{y^2}{144}=\frac{144}{25}\), so the upper point is \(P=\left(13,\frac{144}{5}\right)\). 6. The distance to the right focus \((13,0)\) is \(\frac{144}{5}\,\text{cm}\). 7. The distance to the left focus \((-13,0)\) is \(\sqrt{26^2+\left(\frac{144}{5}\right)^2}=\frac{194}{5}\,\text{cm}\). 8. Their difference is \(\frac{194}{5}-\frac{144}{5}=10\,\text{cm}=2a\), which verifies the constant-difference property.

a) \(\frac{x^2}{25}-\frac{y^2}{144}=1\); asymptotes \(y=\pm\frac{12}{5}x\) b) \(P=\left(13,\frac{144}{5}\right)\) c) Distances \(\frac{194}{5}\,\text{cm}\) and \(\frac{144}{5}\,\text{cm}\); difference \(10\,\text{cm}\)

- The midpoint of the nearest and farthest axial points is the ellipse center. - Use the star's location to find the center-to-focus distance \(c\). - Apply \(b^2=a^2-c^2\) and then compute \(e=\frac{c}{a}\).

1. The endpoints of the major axis are \((2,0)\) and \((-18,0)\), so the center is their midpoint, \((-8,0)\). 2. The semi-major axis is \(a=10\). 3. The star at the origin is \(8\) units from the center, so \(c=8\). 4. For an ellipse, \(b^2=a^2-c^2=100-64=36\). 5. The orbit equation is \(\frac{(x+8)^2}{100}+\frac{y^2}{36}=1\). 6. The eccentricity is \(e=\frac{c}{a}=\frac{4}{5}\). 7. The other focus is \(8\) units left of the center, at \((-16,0)\).

a) \(\frac{(x+8)^2}{100}+\frac{y^2}{36}=1\) b) Eccentricity: \(\frac{4}{5}\); other focus: \((-16,0)\)

- Translate each candidate position so the reported estimate becomes the origin. - Resolve each translated vector along the stated major-axis and minor-axis directions. - Compare the resulting principal coordinates with the ellipse inequality determined by the displayed semi-axis lengths.

1. Relative to the estimate, use principal coordinates \(x'=(x-20)\cos30^\circ+(y-10)\sin30^\circ\) and \(y'=-(x-20)\sin30^\circ+(y-10)\cos30^\circ\). 2. From the diagram, the uncertainty region is \(\frac{x'^2}{36}+\frac{y'^2}{4}\le1\). 3. For \(A\), the relative vector is \((3\sqrt{3},3)\), giving \((x',y')=(6,0)\). Thus the ellipse expression equals \(1\), so \(A\) is on the ellipse. 4. For \(B\), the relative vector is \((2\sqrt{3},2)\), giving \((x',y')=(4,0)\). Thus the value is \(\frac{4}{9}<1\), so \(B\) is inside. 5. For \(C\), the relative vector is \(\left(-\frac{3}{2},\frac{3\sqrt{3}}{2}\right)\), giving \((x',y')=(0,3)\). Thus the value is \(\frac{9}{4}>1\), so \(C\) is outside.

a) \(A\) is on the uncertainty ellipse. b) \(B\) is inside the uncertainty ellipse. c) \(C\) is outside the uncertainty ellipse.

- Convert each full lake dimension into a semi-axis length before writing the ellipse equation. - Substitute the ferry route's constant y-coordinate into the shoreline equation. - Compare the distance between the two intersection x-coordinates with the full major-axis length.

1. The semi-major axis is \(a=13\), and the semi-minor axis is \(b=5\). 2. The shoreline is modeled by \(\frac{x^2}{169}+\frac{y^2}{25}=1\). 3. On the ferry route, \(y=3\). Substitute: \(\frac{x^2}{169}+\frac{9}{25}=1\). 4. Then \(\frac{x^2}{169}=\frac{16}{25}\), so \(x=\pm\frac{52}{5}\). 5. The entry and exit points are \(\left(-\frac{52}{5}, 3\right)\) and \(\left(\frac{52}{5}, 3\right)\). 6. The route length across the lake is \(\frac{52}{5}-\left(-\frac{52}{5}\right)=\frac{104}{5}=20.8\,\text{km}\). 7. The maximum east-west width is \(26\,\text{km}\), so the percentage is \(\frac{20.8}{26}\cdot100\%=80\%\).

a) \(\frac{x^2}{169}+\frac{y^2}{25}=1\) b) \(\left(-\frac{52}{5}, 3\right)\) and \(\left(\frac{52}{5}, 3\right)\) c) \(20.8\,\text{km}\), which is \(80\%\) of the maximum east-west width

- Compute each broken path as the sum of two ordinary distances. - Compare the result with the constant focal-distance sum of the ellipse. - Use the comparison to explain why the exact focus location matters.

1. Through \(A=(10, 0)\), the speaker-to-wall distance is \(3\) and the wall-to-listener distance is \(18\), so the total path length is \(21\,\text{m}\). 2. Through \(B=(0, 6)\), the speaker-to-wall distance is \(\sqrt{(7-0)^2+(0-6)^2}=\sqrt{85}\). 3. The wall-to-listener distance is \(\sqrt{(0+8)^2+(6-0)^2}=10\). 4. The second total path length is \(10+\sqrt{85}\,\text{m}\). 5. The difference is \(21-(10+\sqrt{85})=11-\sqrt{85}\approx1.78\,\text{m}\). 6. If the speaker were at the right focus \((8, 0)\), every one-reflection broken path to the left focus would have length \(2a=20\,\text{m}\). The unequal lengths show that moving the speaker off the focus destroys the constant-path property.

a) Through \(A\): \(21\,\text{m}\); through \(B\): \(10+\sqrt{85}\,\text{m}\approx19.22\,\text{m}\) b) \(11-\sqrt{85}\,\text{m}\approx1.78\,\text{m}\) c) The speaker is not at a focus, so the broken path length to the other focus is no longer constant.

- Use \(c^2=a^2+b^2\) to relate the focus to the nearest vertex. - Read the asymptote slopes from the standard horizontal-hyperbola form. - Substitute the given x-coordinate and use symmetry for the two possible y-coordinates.

1. The hyperbola has \(a=8\), \(b=6\), and \(c=\sqrt{a^2+b^2}=10\), so the stated planet location is a focus. 2. The closest point on the right branch to the right focus is the right vertex \((8, 0)\). 3. The closest distance is \(c-a=10-8=2\) thousand kilometers. 4. The asymptotes are \(y=\pm\frac{b}{a}x=\pm\frac34x\). 5. Their direction angles are \(\theta=\pm\arctan\left(\frac34\right)\approx\pm36.9^\circ\). 6. At \(x=20\), \(\frac{400}{64}-\frac{y^2}{36}=1\). 7. Thus \(\frac{y^2}{36}=\frac{21}{4}\), so \(y=\pm3\sqrt{21}\). 8. The distance between the two positions is \(3\sqrt{21}-(-3\sqrt{21})=6\sqrt{21}\) thousand kilometers, approximately \(27.5\) thousand kilometers.

a) \(2\) thousand kilometers b) \(y=\pm\frac34x\); direction angles \(\pm\arctan\left(\frac34\right)\approx\pm36.9^\circ\) c) \(y=\pm3\sqrt{21}\); separation \(6\sqrt{21}\) thousand kilometers \(\approx27.5\) thousand kilometers

- Read the trimming-line x-coordinate from the diagram and use it to find the two symmetric rim points. - Locate the focus from the standard parabola coefficient. - Use a dot product between the two focus-to-rim vectors to find the included angle.

1. Since \(y^2=4px\), the focal parameter is \(p=3\), so the lamp is at \(F=(3,0)\). 2. From the diagram, the trimming line is \(x=12\). Then \(y^2=144\), so the rim points are \((12,12)\) and \((12,-12)\). 3. Each focus-to-rim distance is \(\sqrt{(12-3)^2+12^2}=15\,\text{cm}\). 4. The focus-to-rim vectors are \((9,12)\) and \((9,-12)\). 5. Their dot product is \(-63\), and each vector has length \(15\). 6. If \(\phi\) is the opening angle, then \(\cos\phi=\frac{-63}{225}=-\frac{7}{25}\), so \(\phi=\arccos\left(-\frac{7}{25}\right)\approx106.3^\circ\).

Rim points: \((12,12)\) and \((12,-12)\) Distance from the lamp to either rim point: \(15\,\text{cm}\) Opening angle: \(\arccos\left(-\frac{7}{25}\right)\approx106.3^\circ\)

- Eliminate the height variable by combining the cone and plane equations. - Complete the square before classifying the projected curve. - Read the semi-axis lengths from standard form before computing area.

1. Points on the cut satisfy both equations, so substitute \(z=2+\frac{x}{2}\) into \(z^2=x^2+y^2\). 2. This gives \(\left(2+\frac{x}{2}\right)^2=x^2+y^2\), so \(\frac34x^2-2x+y^2-4=0\). 3. Complete the square in \(x\): \(\frac34\left(x-\frac43\right)^2+y^2=\frac{16}{3}\). 4. Dividing by \(\frac{16}{3}\) gives \(\frac{\left(x-\frac43\right)^2}{\frac{64}{9}}+\frac{y^2}{\frac{16}{3}}=1\). 5. This ellipse has \(-\frac43\le x\le4\), so \(z=2+\frac{x}{2}\ge\frac43>0\) everywhere on the curve. Thus every projected point belongs to the upper cone. 6. The top-view curve is an ellipse centered at \(\left(\frac43, 0\right)\), with horizontal semi-axis \(\frac83\) and vertical semi-axis \(\frac{4}{\sqrt3}\). 7. Its area is \(\pi\left(\frac83\right)\left(\frac{4}{\sqrt3}\right)=\frac{32\pi}{3\sqrt3}=\frac{32\pi\sqrt3}{9}\) square centimeters.

Top-view equation: \(\frac{\left(x-\frac43\right)^2}{\frac{64}{9}}+\frac{y^2}{\frac{16}{3}}=1\) Conic: ellipse Center: \(\left(\frac43, 0\right)\) Semi-axis lengths: \(\frac83\) and \(\frac{4}{\sqrt3}\) Area: \(\frac{32\pi\sqrt3}{9}\,\text{cm}^2\)