- Start with the standard form for an ellipse centered at \((h, k)\). - A horizontal major axis places the larger squared semi-axis under the x-term. - Remember that the denominators are the squares of the semi-axis lengths.

1. For a horizontal ellipse centered at \((h, k)\), use \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\). 2. Substitute \(h=-2\), \(k=3\), \(a=7\), and \(b=4\). 3. The standard form is \(\frac{(x+2)^2}{49}+\frac{(y-3)^2}{16}=1\).

\(\frac{(x+2)^2}{49}+\frac{(y-3)^2}{16}=1\)

- Start with the standard form centered at \((h, k)\). - A vertical major axis places the larger denominator under the y-term. - Square each semi-axis length before placing it in the denominator.

1. For a vertical ellipse centered at \((h, k)\), use \(\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1\). 2. Here \(h=1\), \(k=-4\), \(a=7\), and \(b=5\). 3. Squaring the semi-axis lengths gives \(a^2=49\) and \(b^2=25\). 4. The standard form is \(\frac{(x-1)^2}{25}+\frac{(y+4)^2}{49}=1\).

\(\frac{(x-1)^2}{25}+\frac{(y+4)^2}{49}=1\)

- Use the center-to-vertex distance for the semi-major axis. - Use the center-to-co-vertex distance for the semi-minor axis. - A horizontal major axis places the larger denominator under the x-term.

1. The center-to-vertex distance is \(a=7\), so \(a^2=49\). 2. The center-to-co-vertex distance is \(b=4\), so \(b^2=16\). 3. The major axis is horizontal, so use \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\). 4. The standard form is \(\frac{(x+2)^2}{49}+\frac{(y-1)^2}{16}=1\).

\(\frac{(x+2)^2}{49}+\frac{(y-1)^2}{16}=1\)

- Start at the marked center and read the horizontal and vertical extreme points from the grid. - Each semi-axis length is the distance from the center to an extreme point. - The longer of the two semi-axes determines the major-axis orientation.

1. The marked center is \((-1, 2)\). 2. The left and right extremes are \((-5, 2)\) and \((3, 2)\), so the horizontal semi-axis has length \(4\). 3. The top and bottom extremes are \((-1, 4)\) and \((-1, 0)\), so the vertical semi-axis has length \(2\). 4. The longer semi-axis is horizontal, so the major axis is horizontal, with semi-major length \(4\) and semi-minor length \(2\).

Center: \((-1, 2)\) Major-axis orientation: horizontal Semi-major axis: \(4\) Semi-minor axis: \(2\)

- The ellipse is defined by a fixed *sum* of two distances. - Find the two distances from the proposed point, then add them.

1. Each focus is \(\sqrt{3^2+4^2}=5\) units from \(P\). 2. The two distances add to \(10\), so \(P\) satisfies the ellipse's defining property.

Yes. The focal distances are \(5\) and \(5\), and their sum is \(10\).

- Which side of the center are the foci on, and are they inside the ellipse? - Compare the formula for \(c\) in an ellipse with the one for a hyperbola.

1. For an ellipse, \(c^2=a^2-b^2\), not \(a^2+b^2\). 2. Thus \(c=\sqrt{25-16}=3\). 3. Its foci are inside the ellipse, so the center-to-focus distance must be less than the semi-major axis.

\(c=3\). The ellipse uses \(c^2=a^2-b^2\), so its foci are closer to the center than the vertices are.

- Read the horizontal and vertical semi-axis lengths from each equation. - A circle has the same radius in every direction. - Think about what happens to the focus distance when \(a=b\).

1. Equation I has equal denominators, so the distances from the center to its horizontal and vertical endpoints are both \(3\): a circle. 2. Equation II has semi-axes \(3\) and \(2\): a noncircular ellipse. 3. When \(a=b\), the focus distance \(c=\sqrt{a^2-b^2}\) becomes \(0\), so the foci meet at the center.

I is a circle; II is a noncircular ellipse. With equal semi-axes, the two foci coincide at the center.

- Use the center-to-vertex distance for the semi-major axis and the center-to-focus distance for the focal distance. - For an ellipse, remember \(a^2=b^2+c^2\), not the hyperbola relationship. - A vertical major axis places \(a^2\) under the y-term.

1. The distance from the center to either vertex is \(a=8\), so \(a^2=64\). 2. The distance from the center to either focus is \(c=6\), so \(c^2=36\). 3. For an ellipse, \(a^2=b^2+c^2\). Thus \(b^2=64-36=28\). 4. Because the major axis is vertical, the standard form is \(\frac{(x-4)^2}{28}+\frac{(y+1)^2}{64}=1\).

\(\frac{(x-4)^2}{28}+\frac{(y+1)^2}{64}=1\)

- Group the terms in each variable before completing the squares. - After completing both squares, divide so the right side is \(1\). - The larger denominator identifies the major-axis direction and the center-to-vertex distance.

1. Group the variable terms: \(4(x^2-6x)+25(y^2+4y)+36=0\). 2. Complete both squares: \(4((x-3)^2-9)+25((y+2)^2-4)+36=0\). 3. Simplify to \(4(x-3)^2+25(y+2)^2=100\). 4. Divide by \(100\): \(\frac{(x-3)^2}{25}+\frac{(y+2)^2}{4}=1\). 5. The center is \((3, -2)\). The major axis is horizontal with \(a=5\), so the vertices are \((-2, -2)\) and \((8, -2)\).

Standard form: \(\frac{(x-3)^2}{25}+\frac{(y+2)^2}{4}=1\) Center: \((3, -2)\) Vertices: \((-2, -2)\) and \((8, -2)\)

- Read each semi-axis length from the corresponding pair of axis intercepts. - Square the semi-axis lengths to form the denominators in standard form. - For an ellipse, use \(a^2=b^2+c^2\) to find the focal distance.

1. The x-intercepts are \((\pm6, 0)\), so the horizontal semi-major axis is \(a=6\). 2. The y-intercepts are \((0, \pm4)\), so the vertical semi-minor axis is \(b=4\). 3. The standard form is \(\frac{x^2}{36}+\frac{y^2}{16}=1\). 4. For an ellipse, \(c^2=a^2-b^2=36-16=20\), so \(c=2\sqrt{5}\). 5. The foci are \((\pm2\sqrt{5}, 0)\).

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

- Group the x-terms and y-terms before completing each square. - Divide by the resulting positive constant to obtain standard form. - For an ellipse, use \(a^2=b^2+c^2\) to find the focal distance.

1. Group the variable terms: \(9(x^2+6x)+16(y^2-4y)+1=0\). 2. Complete both squares: \(9((x+3)^2-9)+16((y-2)^2-4)+1=0\). 3. Simplify to \(9(x+3)^2+16(y-2)^2=144\). 4. Divide by \(144\): \(\frac{(x+3)^2}{16}+\frac{(y-2)^2}{9}=1\). 5. The center is \((-3, 2)\). Here \(a^2=16\), \(b^2=9\), so \(c^2=a^2-b^2=7\). 6. The foci are \((-3-\sqrt{7}, 2)\) and \((-3+\sqrt{7}, 2)\).

Standard form: \(\frac{(x+3)^2}{16}+\frac{(y-2)^2}{9}=1\) Center: \((-3, 2)\) Foci: \((-3-\sqrt{7}, 2)\) and \((-3+\sqrt{7}, 2)\)

- Use the center-to-vertex distance for \(a\) and the center-to-focus distance for \(c\). - For an ellipse, remember \(a^2=b^2+c^2\). - A horizontal major axis places \(a^2\) under the x-term.

1. The center-to-vertex distance is \(a=6\), so \(a^2=36\). 2. The center-to-focus distance is \(c=4\), so \(c^2=16\). 3. For an ellipse, \(a^2=b^2+c^2\). Thus \(b^2=36-16=20\). 4. The major axis is horizontal, so the standard form is \(\frac{(x+4)^2}{36}+\frac{(y-1)^2}{20}=1\).

\(\frac{(x+4)^2}{36}+\frac{(y-1)^2}{20}=1\)

- Use the center-to-vertex distance for \(a\) and the center-to-focus distance for \(c\). - For an ellipse, calculate \(b^2=a^2-c^2\). - A vertical major axis places \(a^2\) under the y-term.

1. The center-to-vertex distance is \(a=7\), so \(a^2=49\). 2. The center-to-focus distance is \(c=3\), so \(c^2=9\). 3. For an ellipse, \(a^2=b^2+c^2\). Thus \(b^2=49-9=40\). 4. Because the major axis is vertical, the standard form is \(\frac{(x+1)^2}{40}+\frac{(y-2)^2}{49}=1\).

\(\frac{(x+1)^2}{40}+\frac{(y-2)^2}{49}=1\)

- Use the center-to-vertex distance to determine the horizontal semi-major axis. - Leave the other denominator unknown in standard form. - The second given point is not on either principal axis, so both squared terms contribute when you substitute it.

1. The distance from the center \((5,-2)\) to the vertex \((13,-2)\) is \(a=8\), so \(a^2=64\). 2. Write \(\frac{(x-5)^2}{64}+\frac{(y+2)^2}{b^2}=1\). 3. Substitute \((9,4)\): \(\frac{16}{64}+\frac{36}{b^2}=1\). 4. Thus \(\frac{36}{b^2}=\frac{3}{4}\), so \(b^2=48\). 5. Since \(64>48\), the horizontal-major-axis condition holds. The equation is \(\frac{(x-5)^2}{64}+\frac{(y+2)^2}{48}=1\).

\(\frac{(x-5)^2}{64}+\frac{(y+2)^2}{48}=1\)

- Group the x-terms and y-terms before completing the squares. - Divide by the resulting positive constant to make the right side \(1\). - The larger denominator identifies the major-axis direction and vertex distance.

1. Group the variable terms: \(25(x^2+4x)+4(y^2-4y)-284=0\). 2. Complete both squares: \(25((x+2)^2-4)+4((y-2)^2-4)-284=0\). 3. Simplify to \(25(x+2)^2+4(y-2)^2=400\). 4. Divide by \(400\): \(\frac{(x+2)^2}{16}+\frac{(y-2)^2}{100}=1\). 5. The center is \((-2, 2)\). The major axis is vertical with \(a=10\), so the vertices are \((-2, 12)\) and \((-2, -8)\).

Standard form: \(\frac{(x+2)^2}{16}+\frac{(y-2)^2}{100}=1\) Center: \((-2, 2)\) Vertices: \((-2, 12)\) and \((-2, -8)\)

- Convert the full major-axis length into a semi-major axis. - Use \(A=\pi ab\) to determine the semi-minor axis. - Place the larger denominator under the horizontal squared term.

1. The major-axis length is \(16\), so the semi-major axis is \(a=8\). 2. The area of an ellipse is \(\pi ab\). Thus \(\pi(8)b=48\pi\), so \(b=6\). 3. The center is \((h,k)=(3,-1)\), and the major axis is horizontal. 4. Therefore the standard equation is \(\frac{(x-3)^2}{64}+\frac{(y+1)^2}{36}=1\).

\(\frac{(x-3)^2}{64}+\frac{(y+1)^2}{36}=1\)

- The midpoint of the foci is the center, and the line through the foci gives the major-axis direction. - Half the major-axis length is \(a\), while the center-to-focus distance is \(c\). - For an ellipse, use \(a^2=b^2+c^2\) before writing standard form.

1. The center is the midpoint of the foci, so it is \((2,4)\). 2. The major axis is horizontal because the foci have the same y-coordinate. 3. The focal distance is \(c=5\), and half the major-axis length is \(a=7\). 4. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=49-25=24\). 5. A horizontal ellipse centered at \((h,k)\) has standard form \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\). 6. Therefore, the equation is \(\frac{(x-2)^2}{49}+\frac{(y-4)^2}{24}=1\).

\(\frac{(x-2)^2}{49}+\frac{(y-4)^2}{24}=1\)

- Read \(a\) from the center-to-vertex distance and \(c\) from the center-to-focus distance. - For an ellipse, remember that \(a^2=b^2+c^2\), not \(c^2=a^2+b^2\). - A vertical major axis places \(a^2\) under the y-term, while the co-vertices lie horizontally from the center.

1. The center-to-vertex distance is \(a=10\), and the center-to-focus distance is \(c=6\). 2. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=100-36=64\) and \(b=8\). 3. Because the major axis is vertical, the standard form is \(\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1\). 4. Substituting \((h,k)=(-4,2)\), \(a^2=100\), and \(b^2=64\) gives \(\frac{(x+4)^2}{64}+\frac{(y-2)^2}{100}=1\). 5. The co-vertices are \(8\) units horizontally from the center, so they are \((-12,2)\) and \((4,2)\).

a) \(\frac{(x+4)^2}{64}+\frac{(y-2)^2}{100}=1\) b) \((-12, 2)\) and \((4, 2)\)

- Read the center and the two semi-axis lengths from the marked points and coordinate grid. - Put the larger semi-axis square under the coordinate that runs along the major axis. - For an ellipse, use \(c^2=a^2-b^2\) to locate the foci from the center.

1. From the graph, the center is \((2,-1)\). 2. The horizontal vertices are \((-4,-1)\) and \((8,-1)\), so the semi-major axis is \(a=6\). 3. The co-vertices are \((2,-5)\) and \((2,3)\), so the semi-minor axis is \(b=4\). 4. The standard equation is \(\frac{(x-2)^2}{36}+\frac{(y+1)^2}{16}=1\). 5. For an ellipse, \(c^2=a^2-b^2=36-16=20\), so \(c=2\sqrt{5}\). 6. The foci lie horizontally from the center at \((2\pm2\sqrt{5},-1)\).

a) \(\frac{(x-2)^2}{36}+\frac{(y+1)^2}{16}=1\) b) \((2-2\sqrt{5}, -1)\) and \((2+2\sqrt{5}, -1)\)

- Read \(c\) from the center-to-focus distance and \(b\) from the center-to-co-vertex distance. - For an ellipse, use \(a^2=b^2+c^2\). - A vertical major axis places \(a^2\) under the y-term and both foci on the vertical line through the center.

1. The center-to-focus distance is \(c=5\). 2. The center-to-co-vertex distance is the semi-minor axis, so \(b=12\). 3. For an ellipse, \(a^2=b^2+c^2=144+25=169\), so \(a=13\). 4. Because the major axis is vertical, the standard form is \(\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1\). 5. Substituting the center and semi-axis lengths gives \(\frac{(x+2)^2}{144}+\frac{(y-1)^2}{169}=1\). 6. The other focus is \(5\) units below the center, at \((-2,-4)\).

Equation: \(\frac{(x+2)^2}{144}+\frac{(y-1)^2}{169}=1\) Other focus: \((-2, -4)\)

- Group the x-terms and y-terms before completing each square. - Normalize the equation so the right side is \(1\), then compare the two denominators. - For an ellipse, use \(c^2=a^2-b^2\) and place the foci along the major axis.

1. Group the variable terms: \(4(x^2-4x)+9(y^2+6y)=-61\). 2. Complete the squares: \(4((x-2)^2-4)+9((y+3)^2-9)=-61\). 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 the larger denominator is under the x-term, so the major axis is horizontal. 6. Here \(a^2=9\), \(b^2=4\), and \(c^2=a^2-b^2=5\). 7. The foci are \((2\pm\sqrt{5},-3)\).

Standard form: \(\frac{(x-2)^2}{9}+\frac{(y+3)^2}{4}=1\) Center: \((2, -3)\) Major axis: horizontal Foci: \((2-\sqrt{5}, -3)\) and \((2+\sqrt{5}, -3)\)

- Substitute the known point into the standard-form equation to solve for the unknown denominator. - Use the positive condition on \(b\) after finding \(b^2\). - For an ellipse, calculate the focal distance with \(c^2=a^2-b^2\).

1. Substitute the given point: \(\frac{64}{100}+\frac{324/25}{b^2}=1\). 2. Simplify to \(\frac{16}{25}+\frac{324}{25b^2}=1\). 3. Thus, \(\frac{324}{25b^2}=\frac{9}{25}\), so \(9b^2=324\) and \(b=6\). 4. The ellipse is \(\frac{x^2}{100}+\frac{y^2}{36}=1\). 5. Here \(a^2=100\) and \(b^2=36\), so \(c^2=a^2-b^2=64\) and \(c=8\). 6. The foci are \((-8,0)\) and \((8,0)\).

\(b=6\) Equation: \(\frac{x^2}{100}+\frac{y^2}{36}=1\) Foci: \((-8, 0)\) and \((8, 0)\)

- Read \(a\) from the center-to-vertex distance and \(c\) from the center-to-focus distance. - For an ellipse, use \(a^2=b^2+c^2\). - The co-vertices lie \(b\) units from the center perpendicular to the major axis.

1. The center-to-vertex distance is \(a=13\), and the center-to-focus distance is \(c=12\). 2. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=169-144=25\) and \(b=5\). 3. Since the major axis is horizontal, the equation is \(\frac{(x-2)^2}{169}+\frac{(y-3)^2}{25}=1\). 4. The co-vertices are \(5\) units above and below the center: \((2,8)\) and \((2,-2)\).

a) \(\frac{(x-2)^2}{169}+\frac{(y-3)^2}{25}=1\) b) \((2, 8)\) and \((2, -2)\)

- Divide each full axis length by \(2\) to obtain the semi-axis lengths. - A vertical major axis places the larger denominator under the y-term. - For an ellipse, use \(c^2=a^2-b^2\) and place the foci along the major axis.

1. Half the major-axis length is \(a=9\), and half the minor-axis length is \(b=6\). 2. Because the major axis is vertical, the standard equation is \(\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1\). 3. Substituting the center and semi-axis lengths gives \(\frac{(x+3)^2}{36}+\frac{(y-2)^2}{81}=1\). 4. For an ellipse, \(c^2=a^2-b^2=81-36=45\), so \(c=3\sqrt{5}\). 5. The foci are \((-3,2\pm3\sqrt{5})\).

Equation: \(\frac{(x+3)^2}{36}+\frac{(y-2)^2}{81}=1\) Foci: \((-3, 2-3\sqrt{5})\) and \((-3, 2+3\sqrt{5})\)

- Half the major-axis length is \(a\), and eccentricity means \(e=\frac{c}{a}\). - For an ellipse, use \(a^2=b^2+c^2\) to find the other denominator. - Place the foci along the horizontal major axis through the center.

1. Half the major-axis length is \(a=13\). 2. The eccentricity is \(e=\frac{c}{a}=\frac{5}{13}\), so \(c=5\). 3. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=169-25=144\). 4. With a horizontal major axis, the equation is \(\frac{(x-3)^2}{169}+\frac{(y+2)^2}{144}=1\). 5. The foci are \(5\) units left and right of the center: \((-2,-2)\) and \((8,-2)\).

Equation: \(\frac{(x-3)^2}{169}+\frac{(y+2)^2}{144}=1\) Foci: \((-2, -2)\) and \((8, -2)\)

- Use \(e=\frac{c}{a}\) to determine the focal distance. - For an ellipse, use \(a^2=b^2+c^2\). - A horizontal major axis places the foci to the left and right of the center.

1. The semi-major axis is \(a=7\). 2. Since \(e=\frac{c}{a}=\frac37\), the focal distance is \(c=3\). 3. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=49-9=40\). 4. The standard equation is \(\frac{(x+5)^2}{49}+\frac{(y-4)^2}{40}=1\). 5. The foci lie \(3\) units horizontally from the center, at \((-8,4)\) and \((-2,4)\).

Equation: \(\frac{(x+5)^2}{49}+\frac{(y-4)^2}{40}=1\) Foci: \((-8,4)\) and \((-2,4)\)

- Half the major-axis length is \(a\), and the center-to-focus distance is \(c\). - For an ellipse, calculate \(b^2=a^2-c^2\). - Place the vertices on the major axis and the co-vertices on the perpendicular axis.

1. Half the major-axis length is \(a=9\), and the focal distance is \(c=7\). 2. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=81-49=32\). 3. The major axis is horizontal, so the equation is \(\frac{x^2}{81}+\frac{y^2}{32}=1\). 4. The vertices are \((-9,0)\) and \((9,0)\). 5. The co-vertices are \((0,-4\sqrt{2})\) and \((0,4\sqrt{2})\).

Equation: \(\frac{x^2}{81}+\frac{y^2}{32}=1\) Vertices: \((-9, 0)\) and \((9, 0)\) Co-vertices: \((0, -4\sqrt{2})\) and \((0, 4\sqrt{2})\)

- Measure \(a\) and \(c\) from the center using the given vertex and focus. - For an ellipse, use \(a^2=b^2+c^2\), not the hyperbola relationship. - Place the larger denominator under the squared expression parallel to the major axis.

1. The distance from the center \((2, -1)\) to the vertex \((9, -1)\) is \(a=7\). 2. The distance from the center to the focus \((7, -1)\) is \(c=5\). 3. For an ellipse, \(a^2=b^2+c^2\), so \(b^2=49-25=24\). 4. Because the major axis is horizontal, the standard equation is \(\frac{(x-2)^2}{49}+\frac{(y+1)^2}{24}=1\). 5. The other vertex is \((2-7, -1)=(-5, -1)\). 6. The co-vertices are \((2, -1\pm\sqrt{24})=(2, -1\pm2\sqrt6)\). 7. The other focus is \((2-5, -1)=(-3, -1)\).

Equation: \(\frac{(x-2)^2}{49}+\frac{(y+1)^2}{24}=1\) Other vertex: \((-5, -1)\) Co-vertices: \((2, -1+2\sqrt6)\) and \((2, -1-2\sqrt6)\) Other focus: \((-3, -1)\)

- Read how far the ellipse extends horizontally and vertically from its center. - Recall what the denominators represent in ellipse standard form. - Check whether the larger denominator is paired with the visibly longer semi-axis.

1. From the graph, the horizontal semi-axis extends \(2\) units from the center, while the vertical semi-axis extends \(5\) units. 2. Standard-form denominators are the squares of the semi-axis lengths, not the lengths themselves. 3. Therefore the denominators should be \(2^2=4\) and \(5^2=25\). 4. The correct equation is \(\frac{(x-2)^2}{4}+\frac{(y+1)^2}{25}=1\).

No. Correct equation: \(\frac{(x-2)^2}{4}+\frac{(y+1)^2}{25}=1\) Marek used the semi-axis lengths as denominators instead of their squares.

- First locate the focus using the two semi-axis lengths. - Because this chord is perpendicular to the major axis, every point on it has the same x-coordinate. - Subtract the y-coordinates of the endpoints to obtain the length.

1. Here \(a=5\), \(b=3\), and \(c=\sqrt{25-9}=4\), so the right focus has \(x=4\). 2. Substitute \(x=4\): \(\frac{16}{25}+\frac{y^2}{9}=1\), giving \(y^2=\frac{81}{25}\). 3. The endpoints are \((4,\frac{9}{5})\) and \((4,-\frac{9}{5})\). The chord length is \(\frac{18}{5}\).

Endpoints: \((4,\frac{9}{5})\), \((4,-\frac{9}{5})\) Length: \(\frac{18}{5}\)

- For a horizontal ellipse, the foci and directrices are positioned along the x-axis. - Find the eccentricity from the center-to-focus and center-to-vertex distances. - Check that the chosen directrix is farther from the center than the right vertex.

1. The semi-major axis is \(a=10\); the focal distance is \(c=\sqrt{100-36}=8\). 2. Thus \(e=\frac{c}{a}=\frac{4}{5}\). 3. The right directrix is \(x=\frac{a}{e}=\frac{25}{2}\), beyond the right vertex.

Eccentricity: \(e=\frac{4}{5}\) Right directrix: \(x=\frac{25}{2}\)

- Read the center-to-vertex distance along the major axis. - Then read the center-to-focus distance. - Eccentricity compares those two distances.

1. The vertices are \((-5,0)\) and \((5,0)\), so \(a=5\). 2. The foci are \((-3,0)\) and \((3,0)\), so \(c=3\). 3. Therefore \(e=\frac{c}{a}=\frac{3}{5}\). Each focus is inside its nearest vertex, \(3\) of the \(5\) units from the center.

\(e=\frac{3}{5}\). The foci lie between the center and the vertices.

- Begin with an ellipse equation centered at the given point. - Clear the denominators before expanding the squares. - Keep every constant term when moving everything to one side.

1. The standard equation is \(\frac{(x-1)^2}{9}+\frac{(y+2)^2}{4}=1\). 2. Multiply by \(36\): \(4(x-1)^2+9(y+2)^2=36\). 3. Expand and collect terms: \(4x^2+9y^2-8x+36y+4=0\).

\(4x^2+9y^2-8x+36y+4=0\)

- The two points give different right triangles when you measure the focus distances. - Distance to the vertical directrix is horizontal, even for the upper point. - Compute each ratio independently, then compare them with \(c/a\).

1. The ellipse has \(a=5\), \(b=3\), and \(c=4\), so its eccentricity is \(e=\frac{4}{5}\). 2. At \(P=(5,0)\), the focus distance is \(1\) and the directrix distance is \(\frac{25}{4}-5=\frac{5}{4}\). The ratio is \(\frac{1}{5/4}=\frac{4}{5}\). 3. At \(Q=(0,3)\), the focus distance is \(\sqrt{4^2+3^2}=5\) and the directrix distance is \(\frac{25}{4}\). The ratio is \(\frac{5}{25/4}=\frac{4}{5}\). 4. Both ratios equal \(e\), despite the points having different positions on the ellipse.

At \(P\): \(\frac{1}{5/4}=\frac{4}{5}\). At \(Q\): \(\frac{5}{25/4}=\frac{4}{5}\). Both equal the eccentricity \(e=\frac{4}{5}\).

- At the right vertex, both distances can be measured horizontally. - Treat the directrix position \(d\) as unknown; the eccentricity compares two distances. - At the second point, find the distance to the focus with a right triangle.

1. At \(V=(5,0)\), the distance to the focus is \(5-3=2\). The distance to \(x=d\) is \(d-5\), because \(d>5\). 2. Use \(\frac{2}{d-5}=\frac{3}{5}\). This gives \(d-5=\frac{10}{3}\), so \(d=\frac{25}{3}\). 3. At \(Q=(0,4)\), the focus distance is \(\sqrt{3^2+4^2}=5\) and the distance to the directrix is \(\frac{25}{3}\). 4. Their ratio is \(\frac{5}{25/3}=\frac{3}{5}\), as required.

Right directrix: \(x=\frac{25}{3}\). At \(Q\), the distance ratio is \(\frac{3}{5}\).

- Use the center and focus to determine the focal distance. - Relate the two semi-axis lengths before using the point on the ellipse. - Check whether every algebraic solution is geometrically possible for a horizontal major axis.

1. The focal distance is \(c=5\), so \(c^2=a^2-b^2\) gives \(b^2=a^2-25\). 2. Substituting \((2, 7)\) into the centered ellipse equation gives \(\frac{9}{a^2}+\frac{16}{b^2}=1\). 3. Replacing \(b^2\) with \(a^2-25\) gives \(\frac{9}{a^2}+\frac{16}{a^2-25}=1\). The resulting equation has solutions \(a^2=5\) and \(a^2=45\). 4. A horizontal ellipse must have \(a^2>c^2=25\), so \(a^2=45\) and \(b^2=20\). 5. The standard form is \(\frac{(x+1)^2}{45}+\frac{(y-3)^2}{20}=1\).

\(\frac{(x+1)^2}{45}+\frac{(y-3)^2}{20}=1\)

- Use the eccentricity to relate the focal distance to the semi-major axis. - Combine that relationship with the area to determine both semi-axis lengths. - Place the larger denominator under the variable aligned with the vertical major axis.

1. The eccentricity gives \(\frac{c^2}{a^2}=\frac{5}{9}\). Since \(b^2=a^2-c^2\), it follows that \(b^2=\frac{4}{9}a^2\), so \(b=\frac{2}{3}a\). 2. The area condition is \(\pi ab=48\pi\), so \(a\left(\frac{2}{3}a\right)=48\). Thus \(a^2=72\). 3. Then \(b^2=32\) and \(c^2=a^2-b^2=40\), so \(c=2\sqrt{10}\). 4. Because the major axis is vertical, the standard form is \(\frac{(x-3)^2}{32}+\frac{(y+2)^2}{72}=1\). 5. The foci are \((3, -2+2\sqrt{10})\) and \((3, -2-2\sqrt{10})\).

The ellipse is \(\frac{(x-3)^2}{32}+\frac{(y+2)^2}{72}=1\). Its foci are \((3, -2+2\sqrt{10})\) and \((3, -2-2\sqrt{10})\).

- Use the vertices to determine the semi-major axis. - Relate the given focal chord length to the two semi-axis lengths. - Place the larger denominator under the variable aligned with the major axis.

1. The distance from the center to either vertex is \(a=8\), so \(a^2=64\). 2. The latus rectum length of an ellipse is \(\frac{2b^2}{a}\). Thus \(\frac{2b^2}{8}=6\), giving \(b^2=24\). 3. Because the major axis is vertical, the equation is \(\frac{(x-2)^2}{24}+\frac{(y+1)^2}{64}=1\). 4. The focal distance satisfies \(c^2=a^2-b^2=64-24=40\), so \(c=2\sqrt{10}\). 5. The foci are \((2, -1+2\sqrt{10})\) and \((2, -1-2\sqrt{10})\).

\(\frac{(x-2)^2}{24}+\frac{(y+1)^2}{64}=1\) Foci: \((2, -1+2\sqrt{10})\) and \((2, -1-2\sqrt{10})\)

- Express all three sides of the focal triangle in terms of ellipse parameters. - Use the equilateral condition before applying the area information. - Recover the remaining semi-axis and focal distance from their standard relationship.

1. The distance from a co-vertex to either focus is \(\sqrt{b^2+c^2}=a\). 2. The base of the focal triangle has length \(2c\). For an equilateral triangle, \(a=2c\), so \(c=\frac{a}{2}\). 3. Then \(b^2=a^2-c^2=\frac{3a^2}{4}\), so \(b=\frac{\sqrt{3}}{2}a\). 4. The area gives \(\pi ab=\frac{\sqrt{3}}{2}\pi a^2=12\sqrt{3}\pi\), so \(a^2=24\). 5. Thus \(b^2=18\) and \(c^2=6\). 6. The equation is \(\frac{x^2}{24}+\frac{y^2}{18}=1\), and the foci are \((\sqrt{6},0)\) and \((-\sqrt{6},0)\).

\(\frac{x^2}{24}+\frac{y^2}{18}=1\) Foci: \((\sqrt{6},0)\) and \((-\sqrt{6},0)\)

- Invert the coordinate transformation before substituting into the circle equation. - Read the ellipse’s geometric features from the resulting standard form. - Apply the transformation directly to the specified point.

1. Solve for the original coordinates: \(u=\frac{x-1}{4}\) and \(v=\frac{y+3}{2}\). 2. Substitution into the unit circle gives \(\frac{(x-1)^2}{16}+\frac{(y+3)^2}{4}=1\). 3. The center is \((1, -3)\), with \(a=4\) and \(b=2\). 4. The vertices are \((-3, -3)\) and \((5, -3)\). The co-vertices are \((1, -5)\) and \((1, -1)\). 5. Since \(c^2=a^2-b^2=12\), the foci are \((1\pm2\sqrt{3}, -3)\). 6. The given circle point maps to \((4(\frac{3}{5})+1, 2(\frac{4}{5})-3)=(\frac{17}{5}, -\frac{7}{5})\).

Equation: \(\frac{(x-1)^2}{16}+\frac{(y+3)^2}{4}=1\) Center: \((1, -3)\) Vertices: \((-3, -3)\), \((5, -3)\) Co-vertices: \((1, -5)\), \((1, -1)\) Foci: \((1\pm2\sqrt{3}, -3)\) Image point: \((\frac{17}{5}, -\frac{7}{5})\)

- Treat the reciprocals of the unknown denominator values as variables. - Each given point creates a linear equation in those reciprocals. - Compare the recovered denominators before locating the foci.

1. Write the ellipse as \(\frac{x^2}{A}+\frac{y^2}{B}=1\), where \(A\) and \(B\) are positive. 2. The two points give \(\frac{16}{A}+\frac{1}{B}=1\) and \(\frac{4}{A}+\frac{9}{B}=1\). 3. Let \(u=\frac{1}{A}\) and \(v=\frac{1}{B}\). Then \(16u+v=1\) and \(4u+9v=1\). 4. Solving gives \(u=\frac{2}{35}\) and \(v=\frac{3}{35}\), so \(A=\frac{35}{2}\) and \(B=\frac{35}{3}\). 5. Since \(A>B\), the major axis is horizontal. 6. The focal distance satisfies \(c^2=A-B=\frac{35}{6}\), so the foci are \((\pm\sqrt{\frac{35}{6}}, 0)\).

\(\frac{x^2}{35/2}+\frac{y^2}{35/3}=1\) Major axis: horizontal Foci: \((\sqrt{\frac{35}{6}}, 0)\), \((-\sqrt{\frac{35}{6}}, 0)\)

- Half the minor-axis length is \(b\), and eccentricity means \(e=\frac{c}{a}\). - For an ellipse, use \(a^2=b^2+c^2\); do not use the hyperbola relation. - A vertical major axis places \(a^2\) under the y-term and the foci above and below the center.

1. The minor-axis length is \(16\), so \(b=8\). 2. The eccentricity gives \(\frac{c}{a}=\frac{3}{5}\), so \(c=\frac{3}{5}a\). 3. For an ellipse, \(a^2=b^2+c^2\). Thus, \(a^2=64+\frac{9}{25}a^2\). 4. Therefore, \(\frac{16}{25}a^2=64\), so \(a^2=100\), \(a=10\), and \(c=6\). 5. With a vertical major axis, the equation is \(\frac{(x-1)^2}{64}+\frac{(y+3)^2}{100}=1\). 6. The foci are \(6\) units above and below the center: \((1,3)\) and \((1,-9)\).

Equation: \(\frac{(x-1)^2}{64}+\frac{(y+3)^2}{100}=1\) Foci: \((1, 3)\) and \((1, -9)\)

- Read \(c\) from the focus and translate the area into the product \(ab\). - Combine \(ab=80\) with the ellipse relation \(a^2=b^2+c^2\). - Place \(a^2\) under the x-term because the major axis is horizontal.

1. The center-to-focus distance is \(c=6\). 2. The area condition gives \(\pi ab=80\pi\), so \(ab=80\). 3. For an ellipse, \(a^2=b^2+c^2=b^2+36\). 4. Since \(b=\frac{80}{a}\), substitute to get \(a^2=\frac{6400}{a^2}+36\). 5. Let \(u=a^2\). Then \(u^2-36u-6400=0\), whose positive solution is \(u=100\). 6. Thus, \(a=10\), \(b=8\), and the equation is \(\frac{(x+1)^2}{100}+\frac{(y-2)^2}{64}=1\). 7. The vertices are \((-11,2)\) and \((9,2)\).

Equation: \(\frac{(x+1)^2}{100}+\frac{(y-2)^2}{64}=1\) Vertices: \((-11, 2)\) and \((9, 2)\)

- Use eccentricity to express \(c\) in terms of \(a\), then apply the ellipse relation. - Combine the resulting ratio between \(a\) and \(b\) with the area formula \(\pi ab\). - A vertical major axis places the larger denominator under the y-term.

1. The eccentricity gives \(c=\frac{8}{17}a\). 2. For an ellipse, \(b^2=a^2-c^2=a^2\left(1-\frac{64}{289}\right)=\frac{225}{289}a^2\), so \(b=\frac{15}{17}a\). 3. The area condition is \(\pi ab=255\pi\), so \(a\left(\frac{15}{17}a\right)=255\). 4. Thus, \(a^2=289\), giving \(a=17\), \(b=15\), and \(c=8\). 5. With a vertical major axis, the equation is \(\frac{(x-1)^2}{225}+\frac{(y+2)^2}{289}=1\). 6. The vertices are \((1,15)\) and \((1,-19)\), and the foci are \((1,6)\) and \((1,-10)\).

Equation: \(\frac{(x-1)^2}{225}+\frac{(y+2)^2}{289}=1\) Vertices: \((1, 15)\) and \((1, -19)\) Foci: \((1, 6)\) and \((1, -10)\)

- Compare the diagonals of the quadrilateral formed by the foci and co-vertices. - Connect the semi-axis lengths and focal distance using the ellipse relationship. - Use the given area only after reducing the number of unknowns.

1. Let \(a\) be the semi-major axis, \(b\) the semi-minor axis and \(c\) the focal distance, with \(c^2=a^2-b^2\). 2. The foci are \((\pm c,0)\), and the co-vertices are \((0,\pm b)\). The quadrilateral has perpendicular diagonals of lengths \(2c\) and \(2b\). 3. Equal square diagonals require \(b=c\). 4. Then \(a^2=b^2+c^2=2b^2\), so \(a=\sqrt{2}b\). 5. From \(\pi ab=\sqrt{2}\pi b^2=18\sqrt{2}\pi\), we obtain \(b^2=18\). 6. Thus \(a^2=36\), \(c^2=18\), and the equation is \(\frac{x^2}{36}+\frac{y^2}{18}=1\). 7. The vertices are \((-6,0)\), \((6,0)\); the co-vertices are \((0,-3\sqrt{2})\), \((0,3\sqrt{2})\); and the foci are \((-3\sqrt{2},0)\), \((3\sqrt{2},0)\).

Equation: \(\frac{x^2}{36}+\frac{y^2}{18}=1\) Vertices: \((-6,0)\), \((6,0)\) Co-vertices: \((0,-3\sqrt{2})\), \((0,3\sqrt{2})\) Foci: \((-3\sqrt{2},0)\), \((3\sqrt{2},0)\)

- Treat the reciprocals of the squared semi-axis lengths as the two unknowns. - Substitute each given point to create a linear system in those unknowns. - After finding the denominators, use the ellipse relationship \(a^2=b^2+c^2\) for the foci.

1. Write the ellipse as \(Ax^2+By^2=1\), where \(A=\frac1{a^2}\) and \(B=\frac1{b^2}\). 2. Substituting \(P\) and multiplying by \(25\) gives \(169A+144B=25\). 3. Substituting \(Q\) and multiplying by \(16\) gives \(169A+36B=16\). 4. Subtract the equations: \(108B=9\), so \(B=\frac1{12}\). 5. Then \(169A+3=16\), so \(A=\frac1{13}\). 6. Therefore the ellipse is \(\frac{x^2}{13}+\frac{y^2}{12}=1\). 7. Since \(a^2=13\) and \(b^2=12\), \(c^2=a^2-b^2=1\), so \(c=1\). 8. The vertices are \((\pm\sqrt{13}, 0)\), the co-vertices are \((0,\pm2\sqrt3)\), and the foci are \((\pm1,0)\).

Equation: \(\frac{x^2}{13}+\frac{y^2}{12}=1\) Vertices: \((\pm\sqrt{13}, 0)\) Co-vertices: \((0, \pm2\sqrt3)\) Foci: \((\pm1, 0)\)

- Relate the directrix distance to the semi-major axis and focal distance. - Use the standard relationship among the two semi-axes and the focal distance. - A latus rectum is perpendicular to the major axis and passes through a focus.

1. The focal distance is \(c=4\). 2. For a horizontal ellipse, each directrix is at distance \(\frac{a}{e}=\frac{a^2}{c}\) from the center. 3. Thus \(\frac{a^2}{4}=9\), so \(a^2=36\) and \(a=6\). 4. The relation \(c^2=a^2-b^2\) gives \(b^2=36-16=20\). 5. The standard equation is \(\frac{x^2}{36}+\frac{y^2}{20}=1\). 6. The vertices are \((\pm6, 0)\), and the co-vertices are \((0, \pm2\sqrt5)\). 7. A latus rectum passes through each focus perpendicular to the major axis. At \(x=\pm4\), the ellipse equation gives \(\frac{16}{36}+\frac{y^2}{20}=1\), so \(y=\pm\frac{10}{3}\).

Equation: \(\frac{x^2}{36}+\frac{y^2}{20}=1\) Vertices: \((-6, 0)\), \((6, 0)\) Co-vertices: \((0, -2\sqrt5)\), \((0, 2\sqrt5)\) Latus-rectum endpoints: \(\left(4, \pm\frac{10}{3}\right)\), \(\left(-4, \pm\frac{10}{3}\right)\)

- Add the two distances from the given point to the foci to obtain \(2a\). - Use the center-to-focus distance for \(c\). - For an ellipse, remember that \(a^2=b^2+c^2\).

1. The distance from \(P\) to the right focus is \(PF_2=\frac{39}{8}\). 2. The distance to the left focus is \(PF_1=\sqrt{(5+5)^2+\left(\frac{39}{8}\right)^2}=\sqrt{100+\frac{1521}{64}}=\frac{89}{8}\). 3. The constant focal-distance sum is \(PF_1+PF_2=\frac{89}{8}+\frac{39}{8}=16\). Thus \(2a=16\), so \(a=8\). 4. The focal distance is \(c=5\). For an ellipse, \(a^2=b^2+c^2\), so \(b^2=64-25=39\). 5. The standard equation is \(\frac{x^2}{64}+\frac{y^2}{39}=1\). 6. The vertices are \((\pm8,0)\), and the co-vertices are \((0,\pm\sqrt{39})\).

Equation: \(\frac{x^2}{64}+\frac{y^2}{39}=1\) Vertices: \((\pm8,0)\) Co-vertices: \((0,\pm\sqrt{39})\)

- Express the two focal distances in terms of the point's \(x\)-coordinate. - Use the ellipse equation to rewrite the squared distance from the center. - The condition depends only on squared coordinates, so check the available sign choices at the end.

1. For a point \(P=(x,y)\) on this ellipse, the focal distances are \(PF_1=13+\frac{5}{13}x\) and \(PF_2=13-\frac{5}{13}x\). Therefore, \(PF_1\cdot PF_2=169-\frac{25}{169}x^2\). 2. From the ellipse equation, \(y^2=144-\frac{144}{169}x^2\). Hence, \(OP^2=x^2+y^2=144+\frac{25}{169}x^2\). 3. Equating the two expressions gives \(144+\frac{25}{169}x^2=169-\frac{25}{169}x^2\), so \(x^2=\frac{169}{2}\). 4. Substitution into the ellipse equation gives \(y^2=72\). 5. Both signs of \(x\) and \(y\) are possible independently, producing four points.

\(\left(\frac{13\sqrt2}{2},6\sqrt2\right)\), \(\left(\frac{13\sqrt2}{2},-6\sqrt2\right)\), \(\left(-\frac{13\sqrt2}{2},6\sqrt2\right)\), and \(\left(-\frac{13\sqrt2}{2},-6\sqrt2\right)\)

- First locate each right focus using the two semi-axis lengths. - At a vertical chord, both endpoints have the same x-coordinate as the focus. - Find the positive and negative y-values from each ellipse equation before comparing the lengths.

1. Ellipse A is \(\frac{x^2}{36}+\frac{y^2}{9}=1\), with right focus at \(x=\sqrt{36-9}=3\sqrt{3}\). 2. At this x-value, \(\frac{y^2}{9}=1-\frac{27}{36}=\frac{1}{4}\). The endpoints have \(y=\pm\frac{3}{2}\), so the chord length is \(3\). 3. Ellipse B is \(\frac{x^2}{36}+\frac{y^2}{16}=1\), with right focus at \(x=\sqrt{36-16}=2\sqrt{5}\). 4. At this x-value, \(\frac{y^2}{16}=1-\frac{20}{36}=\frac{4}{9}\), giving endpoints at \(y=\pm\frac{8}{3}\) and chord length \(\frac{16}{3}\). 5. Ellipse B has the longer chord by \(\frac{16}{3}-3=\frac{7}{3}\).

Ellipse B, by \(\frac{7}{3}\) units.

- What distance total is fixed by the ellipse's major-axis length? - Subtract the two squared distance formulas; see what cancels. - Can the sum and difference of the unknown distances determine each one?

1. The semi-major axis is \(a=5\), so \(PF_L+PF_R=10\). 2. Let \(L=PF_L\) and \(R=PF_R\). Their squared difference is \(L^2-R^2=((x+3)^2+y^2)-((x-3)^2+y^2)=12x\). 3. Since \((L-R)(L+R)=12x\) and \(L+R=10\), it follows that \(L-R=\frac{6x}{5}\). 4. Adding and subtracting the two equations gives \(PF_L=5+\frac{3x}{5}\) and \(PF_R=5-\frac{3x}{5}\).

\(PF_L=5+\frac{3x}{5}\), \(PF_R=5-\frac{3x}{5}\).

- Use the ellipse’s constant sum of focal distances together with the given ratio. - Subtract the two squared-distance equations to eliminate \(y^2\). - Use either focal distance to recover the two symmetric y-values.

1. The foci are \((-5, 0)\) and \((5, 0)\), and the sum of the two focal distances is \(2a=14\). 2. Let \(d_L\) be the distance to the left focus and \(d_R\) the distance to the right focus. The conditions are \(d_R=2d_L\) and \(d_L+d_R=14\). 3. Thus \(d_L=\frac{14}{3}\) and \(d_R=\frac{28}{3}\). 4. Subtracting the squared-distance equations gives \((x-5)^2+y^2-((x+5)^2+y^2)=\frac{784}{9}-\frac{196}{9}\). 5. This simplifies to \(-20x=\frac{196}{3}\), so \(x=-\frac{49}{15}\). 6. Using the left-focus distance gives \((x+5)^2+y^2=\frac{196}{9}\), so \(y^2=\frac{4224}{225}\) and \(y=\pm\frac{8\sqrt{66}}{15}\).

\((-\frac{49}{15}, \frac{8\sqrt{66}}{15})\) and \((-\frac{49}{15}, -\frac{8\sqrt{66}}{15})\)

- Treat the two vector equations as a linear system for the sine and cosine parameters. - Use the unit-circle identity after solving for those parameters. - Read the geometric data from the simplified standard equation.

1. The parameterization gives \(x=6\cos t-8\sin t\) and \(y=4\cos t+3\sin t\). 2. Solve this linear system for the trigonometric values: \(\cos t=\frac{3x+8y}{50}\), \(\sin t=\frac{-4x+6y}{50}\). 3. Use \(\cos^2t+\sin^2t=1\): \((3x+8y)^2+(-4x+6y)^2=2500\). 4. Expanding simplifies to \(25x^2+100y^2=2500\), or \(\frac{x^2}{100}+\frac{y^2}{25}=1\). 5. The semi-axis lengths are \(a=10\) and \(b=5\). Therefore the vertices are \((\pm10, 0)\), the co-vertices are \((0, \pm5)\), and \(c=\sqrt{a^2-b^2}=5\sqrt3\). 6. The foci are \((\pm5\sqrt3, 0)\), and the area is \(\pi ab=50\pi\).

Equation: \(\frac{x^2}{100}+\frac{y^2}{25}=1\) Vertices: \((-10, 0)\), \((10, 0)\) Co-vertices: \((0, -5)\), \((0, 5)\) Foci: \((-5\sqrt3, 0)\), \((5\sqrt3, 0)\) Area: \(50\pi\)