Ellipse Practice Problems with Solutions

We begin with the geometric definition of an ellipse:

$$d(F_1,P)+d(F_2,P)=2a$$

Using the distance formula, we find the distance from each focus to the given point.

Distance from \(F_1(0,-3)\) to \(P(1,4):\)

$$d(F_1,P)=\sqrt{(1-0)^2+(4-(-3))^2}=\sqrt{1^2+(4+3)^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}$$

Distance from \(F_2(0,3)\) to \(P(1,4):\)

$$d(F_2,P)=\sqrt{(1-0)^2+(4-3)^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2}$$

Now, compute 2a:

$$2a=d(F_1,P)+d(F_2,P)=5\sqrt{2}+\sqrt{2}=6\sqrt{2}$$

Therefore:

$$a=\dfrac{6\sqrt{2}}{2}=3\sqrt{2}$$

The center of the ellipse is the midpoint between the foci:

$$\text{Center}=\left(0,\dfrac{-3+3}{2}\right)=(0,0)$$

Because the foci lie on the y-axis, the ellipse is vertically oriented. Its standard form equation is:

$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$

The focal distance 2c equals the distance between the two foci:

$$2c=6 \implies c=3$$

Next, we use the fundamental relationship for an ellipse:

$$c^2=a^2-b^2 \implies b^2=a^2-c^2$$

Since \(a=3\sqrt{2},\) we have:

$$a^2=(3\sqrt{2})^2=9\cdot2=18$$

Substitute \(a^2\) and \(c^2:\)

$$b^2=a^2-c^2=18-9=9$$

Finally, substitute \(a^2=18\) and \(b^2=9\) into the standard equation:

$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$

$$\dfrac{x^2}{9}+\dfrac{y^2}{18}=1$$

We apply the geometric definition of an ellipse:

$$d(F_1, P)+d(F_2,P)=2a$$

Calculate the distance from each focus to point P using the distance formula:

Distance from F₁(-2, 0) to P(3, 1):

$$d(F_1, P)=\sqrt{(3-(-2))^2+(1-0)^2}=\sqrt{(3+2)^2+1^2}=\sqrt{25+1}=\sqrt{26}$$

Distance from F₂(2, 0) to P(3, 1):

$$d(F_2, P)=\sqrt{(3-2)^2+(1-0)^2}=\sqrt{1^2+1^2}=\sqrt{2}$$

Summing both distances yields the value of 2a:

$$2a=d(F_1, P)+d(F_2,P)=\sqrt{26}+\sqrt{2}$$

Next, find the center of the ellipse as the midpoint between the foci, which is the origin (0, 0). Because the foci lie along the x-axis, this is a horizontal ellipse with the following standard form:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

The distance between the foci is 4 units, so:

$$2c=4 \implies c=2$$

Using the relationship between a, b, and c:

$$c^2=a^2-b^2 \implies b^2=a^2-c^2$$

Since \(2a=\sqrt{26}+\sqrt{2},\) we have \(a=\dfrac{\sqrt{26}+\sqrt{2}}{2}.\)

Now calculate \(b^2:\)

$$b^2=a^2-c^2$$

$$=\left(\dfrac{\sqrt{26}+\sqrt{2}}{2}\right)^2-(2)^2$$

$$=\dfrac{(\sqrt{26}+\sqrt{2})^2}{4}-4$$

Substituting \(a^2\) and \(b^2\) gives the standard equation:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

$$\dfrac{x^2}{\dfrac{(\sqrt{26}+\sqrt{2})^2}{4}}+\dfrac{y^2}{\dfrac{(\sqrt{26}+\sqrt{2})^2}{4}-4}=1$$

Simplifying the denominators:

$$\dfrac{x^2}{7+\sqrt{13}}+\dfrac{y^2}{3+\sqrt{13}}=1$$

We use the definition of an ellipse:

$$d(F_1,P)+d(F_2,P)=2a$$

Calculate the distance from each focus to point P(4, 5):

Distance from F₁(2, 3) to P(4, 5):

$$d(F_1,P)=\sqrt{(4-2)^2+(5-3)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$$

Distance from F₂(2, 7) to P(4, 5):

$$d(F_2,P)=\sqrt{(4-2)^2+(5-7)^2}=\sqrt{2^2+(-2)^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$$

Compute 2a:

$$2a=d(F_1,P)+d(F_2,P)=2\sqrt{2}+2\sqrt{2}=4\sqrt{2}$$

Therefore:

$$a=\dfrac{4\sqrt{2}}{2}=2\sqrt{2}$$

The center is the midpoint between the foci:

$$\text{Center}=\left(\dfrac{2+2}{2},\dfrac{3+7}{2}\right)=(2,5)$$

Because the foci share the same x-coordinate (x = 2), the ellipse is oriented vertically. Its standard form equation is:

$$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$$

where the center is \((h, k)=(2, 5).\)

The distance between the foci gives 2c:

$$2c=7-3=4 \implies c=2$$

Using the relationship \(c^2=a^2-b^2 \implies b^2=a^2-c^2:\)

$$a^2=(2\sqrt{2})^2=4\cdot2=8$$

$$c^2=2^2=4$$

Substitute these values to solve for \(b^2:\)

$$b^2=a^2-c^2=8-4=4$$

Substituting \(h=2, k=5, a^2=8,\) and \(b^2=4\) into the standard form yields:

$$\dfrac{(x-2)^2}{4}+\dfrac{(y-5)^2}{8}=1$$

The center is the midpoint between the two foci:

$$\text{Center}=\left(\dfrac{2+2}{2},\dfrac{0+(-6)}{2}\right)=(2, -3)$$

Since the foci share the coordinate x = 2, the major axis is vertical. The standard form is:

$$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$$

where (h, k) = (2, -3).

The major axis has length 10, meaning 2a = 10, which gives \(a=5\) and \(a^2=25.\) The distance between the foci is \(2c=6 \implies c=3 \implies c^2=9.\)

Find \(b^2\) using the characteristic equation:

$$b^2=a^2-c^2$$

$$b^2=25-9$$

$$b^2=16$$

Substitute the center and axis parameters into the standard equation:

$$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$$

$$\dfrac{(x-2)^2}{16}+\dfrac{(y+3)^2}{25}=1$$

The center is the midpoint between the vertices:

$$\text{Center}=\left(\dfrac{-2+4}{2},\dfrac{4+4}{2}\right)=(1,4)$$

Because the vertices vary along the x-coordinate, the major axis is horizontal. The standard form is:

$$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$$

where \((h,k)=(1,4).\)

The distance between the vertices equals the length of the major axis, 2a:

$$2a=\sqrt{(4-(-2))^2+(4-4)^2}=6 \implies a=3 \implies a^2=9$$

The distance from the center to a focus is c. Given one focus at (3, 4):

$$c=\sqrt{(3-1)^2+(4-4)^2}=2 \implies c^2=4$$

We find \(b^2\) using the relationship \(b^2=a^2-c^2:\)

$$b^2=9-4=5$$

Finally, substitute \(h, k, a^2,\) and \(b^2\) into the standard form:

$$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$$

$$\dfrac{(x-1)^2}{9}+\dfrac{(y-4)^2}{5}=1$$

Substitute each point into the general form equation of an ellipse setting A = 1: x² + Cy² + Dx + Ey + F = 0.

For point (0, -1):

$$(0)^2+(-1)^2C+(0)D+(-1)E+F=0 \implies C-E+F=0$$

For point (2, 0):

$$(2)^2+(0)^2C+(2)D+(0)E+F=0 \implies 2D+F=-4$$

For point (4, -1):

$$(4)^2+(-1)^2C+(4)D+(-1)E+F=0 \implies C+4D-E+F=-16$$

For point (2, -2):

$$(2)^2+(-2)^2C+(2)D+(-2)E+F=0 \implies 4C+2D-2E+F=-4$$

This yields a system of four linear equations:

$$\begin{cases}C-E+F=0 \\2D+F=-4 \\C+4D-E+F=-16 \\4C+2D-2E+F=-4 \end{cases}$$

Solving the system gives:

$$C=4, \quad D=-4, \quad E=8, \quad \text{and} \quad F=4$$

Substituting these values back into the general form:

$$x^2+Cy^2+Dx+Ey+F=0$$

produces the equation of the ellipse:

$$x^2+4y^2-4x+8y+4=0$$

Substitute each point into the general equation taking A = 1:

For point (0, 3):

$$(0)^2+(3)^2C+(0)D+(3)E+F=0 \implies 9C+3E+F=0$$

For point (2, 0):

$$(2)^2+(0)^2C+(2)D+(0)E+F=0 \implies 2D+F=-4$$

For point \((1, 3\sqrt{3}/2):\)

$$(1)^2+\left(\dfrac{3\sqrt{3}}{2}\right)^2C+(1)D+\left(\dfrac{3\sqrt{3}}{2}\right)E+F=0 \implies 27C+4D+6\sqrt{3}E+4F=-4$$

For point \((1/2, 3\sqrt{15}/4):\)

$$\left(\dfrac{1}{2}\right)^2+\left(\dfrac{3\sqrt{15}}{4}\right)^2C+\left(\dfrac{1}{2}\right)D+\left(\dfrac{3\sqrt{15}}{4}\right)E+F=0 \implies 135C+8D+12\sqrt{15}E+16F=-4$$

Solving the four-variable linear system gives:

$$C=\dfrac{4}{9}, \quad D=0, \quad E=0, \quad F=-4$$

Substitute these values into the equation:

$$x^2+Cy^2+Dx+Ey+F=0$$

$$x^2+\dfrac{4}{9}y^2-4=0$$

Multiplying the entire equation by 9 yields the general form:

$$9x^2+4y^2-36=0$$

Ellipse A

From the graph, we can observe the following:

• The ellipse is vertically oriented, with foci on the y-axis at (0, -4) and (0, 4).
• The axes of symmetry lie along the coordinate axes.
• The center is at the origin (0, 0).
• A co-vertex is located at (-2, 0), which means the semi-minor axis length is b = 2.
• The focal distance is 2c = 8, so c = 4.

Using the fundamental relationship, find the square of the semi-major axis:

$$c^2=a^2-b^2 \implies a^2=c^2+b^2$$

$$a^2=(4)^2+(2)^2=16+4=20$$

With a center at the origin and a vertical major axis, the standard form is:

$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$

Substituting the values gives:

$$\dfrac{x^2}{4}+\dfrac{y^2}{20}=1$$

Ellipse B

From the graph, we observe:

• The ellipse is horizontally oriented, with foci on the x-axis at (-3, 0) and (3, 0).
• The axes of symmetry lie along the coordinate axes.
• The center is at the origin (0, 0).
• A co-vertex is located at (0, 2), which means the semi-minor axis length is b = 2.
• The focal distance is 2c = 6, so c = 3.

Using the relationship between a, b, and c:

$$a^2=c^2+b^2=3^2+2^2=9+4=13$$

Writing the standard form equation:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

$$\dfrac{x^2}{13}+\dfrac{y^2}{4}=1$$

Given information:

• Center: (0, 0)
• Focus: (3, 0)
• Semi-major axis length: a = 5

Because the center and focus lie on the x-axis, the ellipse is horizontally oriented, giving the standard equation form:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

The distance from the center to the focus gives c = 3. We use the ellipse equation relation to find b:

$$c^2=a^2-b^2 \implies b^2=a^2-c^2$$

$$b^2=5^2-3^2=25-9=16$$

Substituting into standard form:

$$\dfrac{x^2}{25}+\dfrac{y^2}{16}=1$$

Key Elements:

• Foci: (3, 0) and (-3, 0)
• Center: (0, 0)
• Semi-axes: a = 5, b = 4
• Focal distance: 2c = 6
• Vertices: (5, 0) and (-5, 0)
• Co-vertices: (0, 4) and (0, -4)
• Latus rectum (focal width): \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 16}{5}=6.4\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{3}{5}=0.6\)

Given information:

• Center: (0, 0)
• Vertex: (0, 5)
• Co-vertex: (3, 0)
• Major axis along the y-axis

The major axis is along the y-axis, so the standard equation is:

$$\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$$

The vertex at (0, 5) gives a = 5, and the co-vertex at (3, 0) gives b = 3.

Using the ellipse relationship:

$$c^2=a^2-b^2=5^2-3^2=25-9=16 \implies c=4$$

With \(a^2=25\) and \(b^2=9,\) the standard equation is:

$$\dfrac{x^2}{9}+\dfrac{y^2}{25}=1$$

Key Elements:

• Center: (0, 0)
• Foci: (0, 4) and (0, -4)
• Semi-axes: a = 5, b = 3
• Vertices: (0, 5) and (0, -5)
• Co-vertices: (3, 0) and (-3, 0)
• Focal distance: 2c = 8
• Latus rectum (focal width): \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot9}{5}=\dfrac{18}{5}=3.6\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{4}{5}=0.8\)

Given information:

• Vertices: (1, -6) and (9, -6)
• Latus rectum (focal width): 9/2

The vertices lie along a horizontal line (y = -6). The center is the midpoint between the vertices:

$$\text{Center}=\left(\dfrac{1+9}{2},-6\right)=(5,-6)$$

Because the major axis is parallel to the x-axis, the standard form is:

$$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$$

The distance between the center and either vertex gives the semi-major axis length a:

$$a=9-5=4 \implies a^2=16$$

Using the formula for the latus rectum:

$$L_R=\dfrac{2b^2}{a}$$

$$\dfrac{2b^2}{4}=\dfrac{9}{2} \implies b^2=\dfrac{9}{2}\cdot 2=9 \implies b=3$$

Next, determine c:

$$c^2=a^2-b^2=16-9=7 \implies c=\sqrt{7}$$

With \(a^2=16, b^2=9,\) and center at (5, -6), the equation is:

$$\dfrac{(x-5)^2}{16}+\dfrac{(y+6)^2}{9}=1$$

Key Elements:

• Center: (5, -6)
• Foci: \((5-\sqrt{7},-6)\) and \((5+\sqrt{7},-6)\)
• Semi-axes: a = 4, b = 3
• Vertices: (1, -6) and (9, -6)
• Co-vertices: \((5, -3)\) and \((5, -9)\)
• Focal distance: \(2c=2\sqrt{7}\)
• Latus rectum: \(L_R=9/2\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}\approx 0.66\)

Given information:

• Foci: (3, 8) and (3, 2)
• Minor axis length: 2b = 8

The center is the midpoint between the foci:

$$\text{Center}=\left(\dfrac{3+3}{2},\dfrac{8+2}{2}\right)=(3,5)$$

Because the foci share the same x-coordinate, the ellipse is vertically oriented. The standard form is:

$$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$$

The distance between the foci is 2c:

$$2c=8-2=6 \implies c=3$$

Since the minor axis length is 2b = 8, we have \(b=4 \implies b^2=16.\)

Now calculate \(a^2:\)

$$a^2=c^2+b^2=3^2+4^2=9+16=25 \implies a=5$$

Substituting into standard form:

$$\dfrac{(x-3)^2}{16}+\dfrac{(y-5)^2}{25}=1$$

Key Elements:

• Center: (3, 5)
• Foci: (3, 8) and (3, 2)
• Semi-axes: a = 5, b = 4
• Vertices: (3, 10) and (3, 0)
• Co-vertices: (7, 5) and (-1, 5)
• Focal distance: 2c = 6
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 16}{5}=6.4\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{3}{5}=0.6\)

Given information:

• Foci: (2, 4) and (2, 10)
• Vertex: (2, 12)

The center is the midpoint between the foci:

$$\text{Center}=\left(\dfrac{2+2}{2},\dfrac{4+10}{2}\right)=(2,7)$$

The foci share the coordinate x = 2, so the ellipse is vertically oriented:

$$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$$

The distance from the center (2, 7) to the vertex (2, 12) gives a:

$$a=12-7=5 \implies a^2=25$$

The distance between the foci gives 2c:

$$2c=10-4=6 \implies c=3 \implies c^2=9$$

Using \(b^2=a^2-c^2:\)

$$b^2=25-9=16 \implies b=4$$

Substituting into standard form:

$$\dfrac{(x-2)^2}{16}+\dfrac{(y-7)^2}{25}=1$$

Key Elements:

• Center: (2, 7)
• Foci: (2, 10) and (2, 4)
• Semi-axes: a = 5, b = 4
• Vertices: (2, 12) and (2, 2)
• Co-vertices: (6, 7) and (-2, 7)
• Focal distance: 2c = 6
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 16}{5}=6.4\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{3}{5}=0.6\)

Given information:

• Vertices: (-5, 0) and (5, 0)
• Semi-minor axis length: b = 3

The center is the midpoint between the vertices:

$$\text{Center}=\left(\dfrac{-5+5}{2},\dfrac{0+0}{2}\right)=(0,0)$$

The vertices lie along the x-axis, so the ellipse is horizontal:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

The distance from the center to a vertex is a = 5, so \(a^2=25.\)

Compute c using the relation \(c^2=a^2-b^2:\)

$$c^2=25-3^2=25-9=16 \implies c=4$$

Substitute \(a^2=25\) and \(b^2=9\) into the standard equation:

$$\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$$

Key Elements:

• Center: (0, 0)
• Foci: (-4, 0) and (4, 0)
• Semi-axes: a = 5, b = 3
• Vertices: (-5, 0) and (5, 0)
• Co-vertices: (0, 3) and (0, -3)
• Focal distance: 2c = 8
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 9}{5}=3.6\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{4}{5}=0.8\)

Given information:

• Vertices: V(6, 4) and V'(-2, 4)
• Foci: F(5, 4) and F'(-1, 4)

The center is the midpoint between the vertices:

$$\text{Center}=\left(\dfrac{6+(-2)}{2}, \dfrac{4+4}{2}\right)=(2, 4)$$

Both vertices and foci share y = 4, so the ellipse is horizontally oriented:

$$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$$

The distance from the center to a vertex gives a:

$$a=6-2=4 \implies a^2=16$$

The distance from the center to a focus gives c:

$$c=5-2=3 \implies c^2=9$$

Compute \(b^2:\)

$$b^2=a^2-c^2=16-9=7 \implies b=\sqrt{7}$$

The standard equation is:

$$\dfrac{(x-2)^2}{16}+\dfrac{(y-4)^2}{7}=1$$

Key Elements:

• Center: (2, 4)
• Foci: (5, 4) and (-1, 4)
• Semi-axes: \(a=4, b=\sqrt{7}\)
• Vertices: (6, 4) and (-2, 4)
• Co-vertices: \((2, 4+\sqrt{7})\) and \((2, 4-\sqrt{7})\)
• Focal distance: 2c = 6
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 7}{4}=3.5\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{3}{4}=0.75\)

Given information:

• Center: C(7, -2)
• Major axis length: 2a = 8
• Minor axis length: 2b = 4
• Focal axis parallel to the x-axis (horizontal orientation)

Find the semi-axis lengths:

$$a=\dfrac{8}{2}=4$$

$$b=\dfrac{4}{2}=2$$

Calculate c:

$$c^2=a^2-b^2=4^2-2^2=16-4=12 \implies c=\sqrt{12}=2\sqrt{3}$$

The standard equation of this horizontal ellipse is:

$$\dfrac{(x-7)^2}{4^2}+\dfrac{(y+2)^2}{2^2}=1$$

which simplifies to:

$$\dfrac{(x-7)^2}{16}+\dfrac{(y+2)^2}{4}=1$$

Key Elements:

• Center: (7, -2)
• Foci: \((7+2\sqrt{3},-2)\) and \((7-2\sqrt{3},-2)\)
• Semi-axes: a = 4, b = 2
• Vertices: (11, -2) and (3, -2)
• Co-vertices: (7, 0) and (7, -4)
• Focal distance: \(2c=4\sqrt{3}\)
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 4}{4}=2\)
• Eccentricity: \(e=\dfrac{c}{a}=\dfrac{2\sqrt{3}}{4}=\dfrac{\sqrt{3}}{2}\approx 0.866\)

Given information:

• Vertices: (-4, 5) and (16, 5)
• Eccentricity: e = 4/5

The center is the midpoint between the vertices:

$$\text{Center}=\left(\dfrac{-4+16}{2}, \dfrac{5+5}{2}\right)=(6, 5)$$

The vertices share the coordinate y = 5, indicating a horizontal ellipse:

$$\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1$$

The distance from the center to a vertex is a:

$$a=16-6=10 \implies a^2=100$$

Using the eccentricity relation \(e=\dfrac{c}{a}:\)

$$\dfrac{4}{5}=\dfrac{c}{10} \implies c=\dfrac{4}{5}\cdot 10=8$$

Find \(b^2:\)

$$b^2=a^2-c^2=100-8^2=100-64=36 \implies b=6$$

Substituting into standard form:

$$\dfrac{(x-6)^2}{100}+\dfrac{(y-5)^2}{36}=1$$

Key Elements:

• Center: (6, 5)
• Foci: (14, 5) and (-2, 5)
• Semi-axes: a = 10, b = 6
• Vertices: (-4, 5) and (16, 5)
• Co-vertices: (6, 11) and (6, -1)
• Focal distance: 2c = 16
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 36}{10}=7.2\)
• Eccentricity: \(e=4/5=0.8\)

Given information:

• Eccentricity: e = 2/3
• Foci: (0, 0) and (0, -4)

The center is the midpoint between the foci:

$$\text{Center}=\left(\dfrac{0+0}{2}, \dfrac{0+(-4)}{2}\right)=(0,-2)$$

Since the foci lie along the vertical line x = 0, the ellipse is vertically oriented:

$$\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1$$

The distance from the center to either focus gives c = 2. Using \(e=\dfrac{c}{a}:\)

$$\dfrac{2}{3}=\dfrac{2}{a} \implies a=3 \implies a^2=9$$

Now calculate \(b^2:\)

$$b^2=a^2-c^2=9-2^2=9-4=5$$

Substituting into standard form:

$$\dfrac{x^2}{5}+\dfrac{(y+2)^2}{9}=1$$

Key Elements:

• Center: (0, -2)
• Foci: (0, 0) and (0, -4)
• Semi-axes: \(a=3, b=\sqrt{5}\approx 2.24\)
• Vertices: (0, 1) and (0, -5)
• Co-vertices: \((\sqrt{5},-2)\approx (2.24,-2)\) and \((-\sqrt{5},-2)\approx (-2.24,-2)\)
• Focal distance: 2c = 4
• Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 5}{3}=\dfrac{10}{3}\approx 3.33\)
• Eccentricity: \(e=\dfrac{2}{3}\approx 0.67\)

Equation: \(\dfrac{x^2}{16}+\dfrac{y^2}{9}=1\)

The equation is in standard form. The center is at the origin (0, 0), and because the larger denominator lies beneath x², the ellipse is horizontal.

Extract the semi-axis lengths:

$$a^2=16 \implies a=4$$

$$b^2=9 \implies b=3$$

Compute c:

$$c=\sqrt{a^2-b^2}=\sqrt{16-9}=\sqrt{7}$$

Because the major axis is horizontal, the foci lie at a distance of \(\sqrt{7}\) units to the left and right of the center: \((-\sqrt{7}, 0)\) and \((\sqrt{7}, 0).\)

The vertices are located at (-4, 0) and (4, 0), while the co-vertices are at (0, 3) and (0, -3).

Compute the latus rectum and eccentricity:

$$L_R=\dfrac{2b^2}{a}=\dfrac{2\cdot 9}{4}=4.5$$

$$e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}\approx 0.66$$

Key Elements:

• Foci: \((-\sqrt{7}, 0)\) and \((\sqrt{7}, 0)\)
• Center: (0, 0)
• Semi-axes: a = 4, b = 3
• Focal distance: \(2c = 2\sqrt{7}\)
• Vertices: (-4, 0) and (4, 0)
• Co-vertices: (0, 3) and (0, -3)
• Latus rectum: \(L_R=4.5\)
• Eccentricity: \(e\approx 0.66\)

Equation: \(\dfrac{x^2}{4}+\dfrac{y^2}{25}=1\)

The center is (0, 0). The ellipse is vertical because the larger denominator is under \(y^2.\)

Extract the semi-axis values:

$$a^2=25 \implies a=5$$

$$b^2=4 \implies b=2$$

Find the focal distance from the center:

$$c=\sqrt{a^2-b^2}=\sqrt{25-4}=\sqrt{21}$$

Because the major axis is vertical, the foci are located at \((0,-\sqrt{21})\) and \((0, \sqrt{21}).\)

The vertices are at (0, -5) and (0, 5), and the co-vertices are at (-2, 0) and (2, 0).

Calculate the latus rectum and eccentricity:

$$L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 4}{5}=\dfrac{8}{5}=1.6$$

$$e=\dfrac{c}{a}=\dfrac{\sqrt{21}}{5}\approx 0.92$$

Key Elements:

• Foci: \((0,-\sqrt{21})\) and \((0, \sqrt{21})\)
• Center: (0, 0)
• Semi-axes: a = 5, b = 2
• Focal distance: \(2c=2\sqrt{21}\)
• Vertices: (0, -5) and (0, 5)
• Co-vertices: (-2, 0) and (2, 0)
• Latus rectum: \(L_R=1.6\)
• Eccentricity: \(e\approx 0.92\)

Equation: \(\dfrac{(x+3)^2}{36}+\dfrac{(y-4)^2}{16}=1\)

The center is (-3, 4). The ellipse is horizontal because the larger denominator is under \((x+3)^2.\)

Semi-axis lengths:

$$a^2=36 \implies a=6$$

$$b^2=16 \implies b=4$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{36-16}=\sqrt{20}=2\sqrt{5}$$

The foci are located at \((-3-2\sqrt{5}, 4)\) and \((-3+2\sqrt{5}, 4).\)

Vertices: \((-3-6, 4)=(-9, 4)\) and \((-3+6, 4)=(3, 4)\)

Co-vertices: \((-3, 4-4)=(-3, 0)\) and \((-3, 4+4)=(-3, 8)\)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 16}{6}=\dfrac{32}{6}\approx 5.33\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{2\sqrt{5}}{6}=\dfrac{\sqrt{5}}{3}\approx 0.75\)

Key Elements:

• Foci: \((-3-2\sqrt{5}, 4)\) and \((-3+2\sqrt{5}, 4)\)
• Center: (-3, 4)
• Semi-axes: a = 6, b = 4
• Focal distance: \(2c=4\sqrt{5}\)
• Vertices: (-9, 4) and (3, 4)
• Co-vertices: (-3, 0) and (-3, 8)
• Latus rectum: \(L_R\approx 5.33\)
• Eccentricity: \(e\approx 0.75\)

Equation: \(\dfrac{(x-1)^2}{9}+\dfrac{(y+2)^2}{25}=1\)

The center is (1, -2). The ellipse is vertical because the larger denominator is under \((y+2)^2.\)

Semi-axis lengths:

$$a^2=25 \implies a=5$$

$$b^2=9 \implies b=3$$

Focal distance:

$$c=\sqrt{a^2-b^2}=\sqrt{25-9}=\sqrt{16}=4$$

The foci are located at (1, -6) and (1, 2).

Vertices: (1, -7) and (1, 3)

Co-vertices: (-2, -2) and (4, -2)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 9}{5}=3.6\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{4}{5}=0.8\)

Key Elements:

• Foci: (1, -6) and (1, 2)
• Center: (1, -2)
• Semi-axes: a = 5, b = 3
• Focal distance: 2c = 8
• Vertices: (1, -7) and (1, 3)
• Co-vertices: (-2, -2) and (4, -2)
• Latus rectum: \(L_R=3.6\)
• Eccentricity: \(e=0.8\)

Equation: \(\dfrac{(x-1)^2}{36}+\dfrac{(y-\frac{5}{2})^2}{49}=1\)

The center is (1, 5/2). The ellipse is vertical because the larger denominator is under \(\left(y-\dfrac{5}{2}\right)^2.\)

Semi-axis lengths:

$$a^2=49 \implies a=7$$

$$b^2=36 \implies b=6$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{49-36}=\sqrt{13}$$

Foci:

$$\left(1, \dfrac{5}{2}-\sqrt{13}\right) \quad \text{and} \quad \left(1, \dfrac{5}{2}+\sqrt{13}\right)$$

Vertices:

$$\left(1, \dfrac{5}{2}-7\right)=\left(1,-\dfrac{9}{2}\right)$$

$$\left(1, \dfrac{5}{2}+7\right)=\left(1, \dfrac{19}{2}\right)$$

Co-vertices:

$$\left(1-6, \dfrac{5}{2}\right)=\left(-5, \dfrac{5}{2}\right)$$

$$\left(1+6, \dfrac{5}{2}\right)=\left(7, \frac{5}{2}\right)$$

Latus rectum:

$$L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 36}{7}=\dfrac{72}{7}\approx 10.29$$

Eccentricity:

$$e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{7}\approx 0.52$$

Key Elements:

• Foci: \(\left(1, \dfrac{5}{2}-\sqrt{13}\right)\) and \(\left(1, \dfrac{5}{2}+\sqrt{13}\right)\)
• Center: (1, 5/2)
• Semi-axes: a = 7, b = 6
• Focal distance: \(2c=2\sqrt{13}\)
• Vertices: \(\left(1,-\dfrac{9}{2}\right)\) and \(\left(1, \dfrac{19}{2}\right)\)
• Co-vertices: \(\left(-5, \dfrac{5}{2}\right)\) and \(\left(7, \dfrac{5}{2}\right)\)
• Latus rectum: \(L_R\approx 10.29\)
• Eccentricity: \(e\approx 0.52\)

Equation: \(\dfrac{x^2}{25}+\dfrac{(y+5)^2}{16}=1\)

The center is (0, -5). The ellipse is horizontal because the larger denominator is under \(x^2.\)

Semi-axis lengths:

$$a^2=25 \implies a=5$$

$$b^2=16 \implies b=4$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{25-16}=\sqrt{9}=3$$

Foci: (-3, -5) and (3, -5)

Vertices: (-5, -5) and (5, -5)

Co-vertices: (0, -9) and (0, -1)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 16}{5}=6.4\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{3}{5}=0.6\)

Key Elements:

• Foci: (-3, -5) and (3, -5)
• Center: (0, -5)
• Semi-axes: a = 5, b = 4
• Focal distance: 2c = 6
• Vertices: (-5, -5) and (5, -5)
• Co-vertices: (0, -9) and (0, -1)
• Latus rectum: \(L_R=6.4\)
• Eccentricity: e = 0.6

Equation: \(3x^2+4y^2=12\)

This equation is not in standard form. Divide every term by 12:

$$\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$$

The center is (0, 0), and the ellipse is horizontal because the larger denominator is under \(x^2.\)

Semi-axis lengths:

$$a^2=4 \implies a=2$$

$$b^2=3 \implies b=\sqrt{3}$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{4-3}=1$$

Foci: (-1, 0) and (1, 0)

Vertices: (-2, 0) and (2, 0)

Co-vertices: \((0,-\sqrt{3})\) and \((0, \sqrt{3})\)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 3}{2}=3\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{1}{2}=0.5\)

Key Elements:

• Foci: (-1, 0) and (1, 0)
• Center: (0, 0)
• Semi-axes: \(a=2, b=\sqrt{3}\)
• Focal distance: 2c = 2
• Vertices: (-2, 0) and (2, 0)
• Co-vertices: \((0,-\sqrt{3})\) and \((0, \sqrt{3})\)
• Latus rectum: \(L_R=3\)
• Eccentricity: \(e=0.5\)

Equation: \(25x^2+16y^2=400\)

Divide by 400 to obtain the standard form:

$$\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$$

The center is (0, 0). The ellipse is vertical because the larger denominator is under \(y^2.\)

Semi-axis lengths:

$$a^2=25 \implies a=5$$

$$b^2=16 \implies b=4$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{25-16}=\sqrt{9}=3$$

Foci: (0, -3) and (0, 3)

Vertices: (0, -5) and (0, 5)

Co-vertices: (-4, 0) and (4, 0)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 16}{5}=6.4\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{3}{5}=0.6\)

Key Elements:

• Foci: (0, -3) and (0, 3)
• Center: (0, 0)
• Semi-axes: a = 5, b = 4
• Focal distance: 2c = 6
• Vertices: (0, -5) and (0, 5)
• Co-vertices: (-4, 0) and (4, 0)
• Latus rectum: \(L_R=6.4\)
• Eccentricity: e = 0.6

Equation: \(4x^2+9y^2-16x-32=0\)

Completing the square (detailed in the next section) yields the standard equation:

$$\dfrac{(x-2)^2}{12}+\dfrac{y^2}{16/3}=1$$

The center is (2, 0). The ellipse is horizontal because 12 is greater than 16/3.

Semi-axis lengths:

$$a^2=12 \implies a=2\sqrt{3}$$

$$b^2=\dfrac{16}{3} \implies b=\dfrac{4}{\sqrt{3}}=\dfrac{4\sqrt{3}}{3}$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{12-\dfrac{16}{3}}=\sqrt{\dfrac{20}{3}}=\dfrac{2\sqrt{15}}{3}$$

Foci:

$$\left(2-\dfrac{2\sqrt{15}}{3}, 0\right) \quad \text{and} \quad \left(2+\dfrac{2\sqrt{15}}{3}, 0\right)$$

Vertices:

$$(2-2\sqrt{3}, 0) \quad \text{and} \quad (2+2\sqrt{3}, 0)$$

Co-vertices:

$$\left(2,-\dfrac{4\sqrt{3}}{3}\right) \quad \text{and} \quad \left(2, \dfrac{4\sqrt{3}}{3}\right)$$

Latus rectum:

$$L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot \dfrac{16}{3}}{2\sqrt{3}}=\dfrac{16\sqrt{3}}{9}$$

Eccentricity:

$$e=\dfrac{c}{a}=\dfrac{\dfrac{2\sqrt{15}}{3}}{2\sqrt{3}}=\dfrac{\sqrt{5}}{3}\approx 0.75$$

Key Elements:

• Foci: \(\left(2-\dfrac{2\sqrt{15}}{3}, 0\right)\) and \(\left(2+\dfrac{2\sqrt{15}}{3}, 0\right)\)
• Center: (2, 0)
• Semi-axes: \(a=2\sqrt{3}, b=\dfrac{4\sqrt{3}}{3}\)
• Focal distance: \(2c=\dfrac{4\sqrt{15}}{3}\)
• Vertices: \((2-2\sqrt{3}, 0)\) and \((2+2\sqrt{3}, 0)\)
• Co-vertices: \(\left(2,-\dfrac{4\sqrt{3}}{3}\right)\) and \(\left(2, \dfrac{4\sqrt{3}}{3}\right)\)
• Latus rectum: \(L_R=\dfrac{16\sqrt{3}}{9}\)
• Eccentricity: \(e\approx 0.75\)

Equation: \(16x^2+9y^2-64x-54y+1=0\)

Completing the square produces the standard form:

$$\dfrac{(x-2)^2}{9}+\dfrac{(y-3)^2}{16}=1$$

The center is (2, 3). The ellipse is vertical because the larger denominator is under \((y-3)^2.\)

Semi-axis lengths:

$$a^2=16 \implies a=4$$

$$b^2=9 \implies b=3$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{16-9}=\sqrt{7}$$

Foci:

$$(2, 3-\sqrt{7}) \quad \text{and} \quad (2, 3+\sqrt{7})$$

Vertices: (2, -1) and (2, 7)

Co-vertices: (-1, 3) and (5, 3)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 9}{4}=4.5\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}\approx 0.66\)

Key Elements:

• Foci: \((2, 3-\sqrt{7})\) and \((2, 3+\sqrt{7})\)
• Center: (2, 3)
• Semi-axes: a = 4, b = 3
• Focal distance: \(2c=2\sqrt{7}\)
• Vertices: (2, -1) and (2, 7)
• Co-vertices: (-1, 3) and (5, 3)
• Latus rectum: \(L_R=4.5\)
• Eccentricity: \(e\approx 0.66\)

Equation: \(8x^2+25y^2-48x+50y+47=0\)

Completing the square gives the standard equation:

$$\dfrac{(x-3)^2}{25/4}+\dfrac{(y+1)^2}{2}=1$$

The center is (3, -1). Since 25/4 > 2, the ellipse is horizontal.

Semi-axis lengths:

$$a=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}=2.5$$

$$b=\sqrt{2}$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{\dfrac{25}{4}-2}=\sqrt{\dfrac{17}{4}}=\dfrac{\sqrt{17}}{2}$$

Foci:

$$\left(3-\dfrac{\sqrt{17}}{2}, -1\right) \quad \text{and} \quad \left(3+\dfrac{\sqrt{17}}{2}, -1\right)$$

Vertices: (0.5, -1) and (5.5, -1)

Co-vertices: \((3, -1-\sqrt{2})\) and \((3, -1+\sqrt{2})\)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 2}{2.5}=1.6\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{\sqrt{17}/2}{2.5}=\dfrac{\sqrt{17}}{5}\approx 0.82\)

Key Elements:

• Center: (3, -1)
• Foci: \(\left(3-\dfrac{\sqrt{17}}{2}, -1\right)\) and \(\left(3+\dfrac{\sqrt{17}}{2}, -1\right)\)
• Semi-axes: \(a=2.5, b=\sqrt{2}\)
• Focal distance: \(2c=\sqrt{17}\)
• Vertices: (0.5, -1) and (5.5, -1)
• Co-vertices: \((3, -1-\sqrt{2})\) and \((3, -1+\sqrt{2})\)
• Latus rectum: \(L_R=1.6\)
• Eccentricity: \(e\approx 0.82\)

Equation: \(36x^2+25y^2+72x-200y-464=0\)

Completing the square yields the standard form:

$$\dfrac{(x+1)^2}{25}+\dfrac{(y-4)^2}{36}=1$$

The center is (-1, 4). The ellipse is vertical because the larger denominator is under \((y-4)^2.\)

Semi-axis lengths:

$$a=\sqrt{36}=6$$

$$b=\sqrt{25}=5$$

Focal distance from center:

$$c=\sqrt{a^2-b^2}=\sqrt{36-25}=\sqrt{11}$$

Foci: \((-1, 4-\sqrt{11})\) and \((-1, 4+\sqrt{11})\)

Vertices: (-1, -2) and (-1, 10)

Co-vertices: (-6, 4) and (4, 4)

Latus rectum: \(L_R=\dfrac{2b^2}{a}=\dfrac{2 \cdot 25}{6}=\dfrac{50}{6}\approx 8.33\)

Eccentricity: \(e=\dfrac{c}{a}=\dfrac{\sqrt{11}}{6}\approx 0.55\)

Key Elements:

• Center: (-1, 4)
• Foci: \((-1, 4-\sqrt{11})\) and \((-1, 4+\sqrt{11})\)
• Semi-axes: a = 6, b = 5
• Focal distance: \(2c=2\sqrt{11}\)
• Vertices: (-1, -2) and (-1, 10)
• Co-vertices: (-6, 4) and (4, 4)
• Latus rectum: \(L_R\approx 8.33\)
• Eccentricity: \(e\approx 0.55\)

Equation: \(3x^2+4y^2=12\)

This equation contains squared terms in x and y with no linear terms, indicating that the center is at the origin.

In standard form, the right-hand side must equal 1. Divide both sides by 12:

$$\dfrac{3x^2}{12}+\dfrac{4y^2}{12}=\dfrac{12}{12}$$

Simplify each fraction:

$$\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$$

The equation is now in standard form.

Equation: \(25x^2+16y^2=400\)

Because there are no linear terms in x or y, the center is at (0, 0).

Divide the entire equation by 400 so that the right side equals 1:

$$\dfrac{25x^2}{400}+\dfrac{16y^2}{400}=\dfrac{400}{400}$$

Simplify each fraction:

$$\dfrac{x^2}{16}+\dfrac{y^2}{25}=1$$

Equation: \(4x^2+9y^2-16x-32=0\)

There is a linear term in x, which indicates the center is not at the origin, but no linear term in y.

Rearrange the equation and move the constant to the right side:

$$4x^2-16x+9y^2=32$$

Factor out 4 from the x-terms:

$$4(x^2-4x)+9y^2=32$$

Complete the square for the x-terms. Take half the coefficient of x (-4/2 = -2) and square it ((-2)² = 4). Add and subtract 4 inside the brackets:

$$4[(x^2-4x+4)-4]+9y^2=32$$

Write the perfect square trinomial:

$$4[(x-2)^2-4]+9y^2=32$$

Distribute the 4:

$$4(x-2)^2-16+9y^2=32$$

Add 16 to both sides:

$$4(x-2)^2+9y^2=48$$

Divide both sides by 48 to make the right side equal to 1:

$$\dfrac{4(x-2)^2}{48}+\dfrac{9y^2}{48}=\dfrac{48}{48}$$

Simplify each coefficient:

$$\dfrac{(x-2)^2}{12}+\dfrac{y^2}{16/3}=1$$

Equation: \(16x^2+9y^2-64x-54y+1=0\)

This equation has linear terms in both variables, so we must complete the square for both x and y.

Group the x and y terms and move the constant term to the right side:

$$16x^2-64x+9y^2-54y=-1$$

Factor out the leading coefficients from each variable group:

$$16[x^2-4x]+9[y^2-6y]=-1$$

Complete the square:

For x, half of -4 is -2, and (-2)² = 4:

$$16[(x^2-4x+4)-4]=16[(x-2)^2-4]$$

For y, half of -6 is -3, and (-3)² = 9:

$$9[(y^2-6y+9)-9]=9[(y-3)^2-9]$$

Substitute these expressions back into the equation:

$$16[(x-2)^2-4]+9[(y-3)^2-9]=-1$$

Distribute 16 and 9:

$$16(x-2)^2-64+9(y-3)^2-81=-1$$

Add 64 and 81 to the right side:

$$16(x-2)^2+9(y-3)^2=144$$

Divide the entire equation by 144:

$$\dfrac{16(x-2)^2}{144}+\dfrac{9(y-3)^2}{144}=\dfrac{144}{144}$$

Simplify:

$$\dfrac{(x-2)^2}{9}+\dfrac{(y-3)^2}{16}=1$$

Equation: \(8x^2+25y^2-48x+50y+47=0\)

Group the x and y terms and move 47 to the right side:

$$8x^2-48x+25y^2+50y=-47$$

Factor out the leading coefficients:

$$8[x^2-6x]+25[y^2+2y]=-47$$

Complete the square for x (half of -6 squared is 9):

$$8[(x^2-6x+9)-9]=8[(x-3)^2-9]$$

Complete the square for y (half of 2 squared is 1):

$$25[(y^2+2y+1)-1]=25[(y+1)^2-1]$$

Substitute into the equation:

$$8[(x-3)^2-9]+25[(y+1)^2-1]=-47$$

Distribute 8 and 25:

$$8(x-3)^2-72+25(y+1)^2-25=-47$$

Combine and add constant terms to the right side:

$$8(x-3)^2+25(y+1)^2=50$$

Divide through by 50:

$$\dfrac{8(x-3)^2}{50}+\dfrac{25(y+1)^2}{50}=\dfrac{50}{50}$$

Simplify the fractions:

$$\dfrac{(x-3)^2}{25/4}+\dfrac{(y+1)^2}{2}=1$$

Equation: \(36x^2+25y^2+72x-200y-464=0\)

Group terms and move the constant:

$$36x^2+72x+25y^2-200y=464$$

Factor out coefficients:

$$36[x^2+2x]+25[y^2-8y]=464$$

Complete the square for x:

$$36[(x^2+2x+1)-1]=36[(x+1)^2-1]$$

Complete the square for y:

$$25[(y^2-8y+16)-16]=25[(y-4)^2-16]$$

Substitute and distribute:

$$36(x+1)^2-36+25(y-4)^2-400=464$$

Add 36 and 400 to the right side:

$$36(x+1)^2+25(y-4)^2=900$$

Divide by 900:

$$\dfrac{36(x+1)^2}{900}+\dfrac{25(y-4)^2}{900}=\dfrac{900}{900}$$

Simplify coefficients:

$$\dfrac{(x+1)^2}{25}+\dfrac{(y-4)^2}{36}=1$$

Equation: \(2x^2+4x+7y^2-28y-40=0\)

Group terms and move 40 to the right side:

$$2x^2+4x+7y^2-28y=40$$

Factor out leading coefficients:

$$2[x^2+2x]+7[y^2-4y]=40$$

Complete the square for x and y:

$$2[(x^2+2x+1)-1]=2[(x+1)^2-1]$$

$$7[(y^2-4y+4)-4]=7[(y-2)^2-4]$$

Substitute and distribute:

$$2(x+1)^2-2+7(y-2)^2-28=40$$

Add 2 and 28 to the right side:

$$2(x+1)^2+7(y-2)^2=70$$

Divide by 70:

$$\dfrac{2(x+1)^2}{70}+\dfrac{7(y-2)^2}{70}=\dfrac{70}{70}$$

Simplify:

$$\dfrac{(x+1)^2}{35}+\dfrac{(y-2)^2}{10}=1$$

Equation 1

$$\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$$

Multiply both sides by \(16\cdot 9=144\) to clear denominators:

$$144 \cdot \left(\dfrac{x^2}{16}\right)+144 \cdot \left(\dfrac{y^2}{9}\right)=144 \cdot 1$$

Distribute and simplify:

$$9x^2+16y^2=144$$

Subtract 144 to set the equation equal to 0:

$$9x^2+16y^2-144=0$$

This is the general form.

Equation 2

$$\dfrac{x^2}{4}+\dfrac{y^2}{25}=1$$

Multiply through by \(4 \cdot 25=100:\)

$$100 \cdot \left(\dfrac{x^2}{4}\right)+100 \cdot \left(\dfrac{y^2}{25}\right)=100 \cdot 1$$

$$25x^2+4y^2=100$$

Set equal to zero:

$$25x^2+4y^2-100=0$$

Equation 3

$$\dfrac{(x+3)^2}{36}+\dfrac{(y-4)^2}{16}=1$$

Multiply through by \(36 \cdot 16=576:\)

$$16(x+3)^2+36(y-4)^2=576$$

Expand the squared binomials:

$$16(x^2+6x+9)+36(y^2-8y+16)=576$$

$$16x^2+96x+144+36y^2-288y+576=576$$

Rearrange into general form:

$$16x^2+36y^2+96x-288y+144+576-576=0$$

$$16x^2+36y^2+96x-288y+144=0$$

Simplify by dividing by 4:

$$4x^2+9y^2+24x-72y+36=0$$

Equation 4

$$\dfrac{(x-1)^2}{9}+\dfrac{(y+2)^2}{25}=1$$

Multiply through by \(9 \cdot 25=225:\)

$$25(x-1)^2+9(y+2)^2=225$$

Expand both binomials:

$$25(x^2-2x+1)+9(y^2+4y+4)=225$$

$$25x^2-50x+25+9y^2+36y+36=225$$

Rearrange and combine constants:

$$25x^2+9y^2-50x+36y+25+36-225=0$$

$$25x^2+9y^2-50x+36y-164=0$$

$$\dfrac{(x-1)^2}{36}+\dfrac{(y-\frac{5}{2})^2}{49}=1$$

Multiply through by \(36 \cdot 49=1764:\)

$$49(x-1)^2+36\left(y-\dfrac{5}{2}\right)^2=1764$$

Expand each squared term:

$$49(x^2-2x+1)+36\left(y^2-5y+\dfrac{25}{4}\right)=1764$$

$$49x^2-98x+49+36y^2-180y+225=1764$$

Combine like terms:

$$49x^2+36y^2-98x-180y+49+225-1764=0$$

$$49x^2+36y^2-98x-180y-1490=0$$

$$\dfrac{x^2}{25}+\dfrac{(y+5)^2}{16}=1$$

Multiply through by \(25 \cdot 16=400:\)

$$16x^2+25(y+5)^2=400$$

Expand the binomial:

$$16x^2+25(y^2+10y+25)=400$$

$$16x^2+25y^2+250y+625=400$$

Rearrange into general form:

$$16x^2+25y^2+250y+625-400=0$$

$$16x^2+25y^2+250y+225=0$$

Given Data

• Minimum distance (perigee): d_min = 356,000 km
• Maximum distance (apogee): d_max = 406,000 km

In an elliptical orbit:

• The major axis length (2a) is the sum of the perigee and apogee distances.
• The focal distance (2c) is the distance between the two foci.
• The eccentricity is defined as \(e=\dfrac{c}{a}.\)

The total length of the major axis is:

$$2a=d_{\text{min}}+d_{\text{max}}$$

$$2a=356000+406000=762000 \ \text{km}$$

Therefore, the semi-major axis length is:

$$a=\dfrac{2a}{2}=\dfrac{762000}{2}=381000 \ \text{km}$$

The focal length c is given by:

$$c=a-d_{\text{min}}$$

$$c=381000-356000=25000 \ \text{km}$$

Compute the eccentricity:

$$e=\dfrac{c}{a}$$

$$e=\dfrac{25000}{381000}\approx 0.0656$$

Calculate the semi-minor axis length b:

$$b=\sqrt{a^2-c^2}$$

$$b=\sqrt{(381000)^2-(25000)^2}$$

$$b=\sqrt{145161000000-625000000}=\sqrt{144536000000}\approx 380179 \ \text{km}$$

Final Results:

• Minor axis length: \(2b \approx 760358 \ \text{km}\)
• Eccentricity: \(e \approx 0.0656\)
• Major axis length: \(2a = 762000 \ \text{km}\)

The semi-major axis is a = 0.387 AU and the eccentricity is \(e=\dfrac{c}{a}=0.206:\)

$$\dfrac{c}{0.387}=0.206 \implies c=0.079722 \ \text{AU}$$

Using \(b=\sqrt{a^2-c^2}:\)

$$b=\sqrt{(0.387)^2-(0.079722)^2}\approx 0.3787 \ \text{AU}$$

Therefore, the semi-minor axis length is approximately 0.3787 AU.

Given information:

• Perihelion: d_min = 8.29 AU
• Aphelion: d_max = 10.03 AU

Centering the coordinate system at the center of the ellipse, with the major axis along the x-axis, the standard equation is:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

where:

• a is the semi-major axis length.
• b is the semi-minor axis length.
• c is the focal distance, with \(c=\sqrt{a^2-b^2}.\)

The major axis length (2a) is the sum of perihelion and aphelion:

$$2a=d_{\text{min}}+d_{\text{max}}=8.29+10.03=18.32 \ \text{AU}$$

Therefore, the semi-major axis is:

$$a=\dfrac{18.32}{2}=9.16 \ \text{AU}$$

The focal distance c is:

$$c=a-d_{\text{min}}=9.16-8.29=0.87 \ \text{AU}$$

Calculate the semi-minor axis b:

$$b=\sqrt{a^2-c^2}=\sqrt{(9.16)^2-(0.87)^2}=\sqrt{83.9056-0.7569}=\sqrt{83.1487}\approx 9.12 \ \text{AU}$$

With a = 9.16 and b ≈ 9.12, the standard equation centered at (0, 0) is:

$$\dfrac{x^2}{9.16^2}+\dfrac{y^2}{9.12^2}=1$$

$$\dfrac{x^2}{83.91}+\dfrac{y^2}{83.17}=1$$

This equation models Saturn's orbit around the Sun in astronomical units (AU).

Given information:

• Width of the base: 2a = 20 m, so a = 10 m.
• Maximum height at the center: b = 6 m (semi-minor vertical axis).

Placing the center at the origin (0, 0) with a horizontal major axis, the equation of the full ellipse is:

$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$

Because the arch is the upper half of the ellipse, we only consider the positive values of y.

Substituting a = 10 and b = 6:

$$\dfrac{x^2}{10^2}+\dfrac{y^2}{6^2}=1$$

$$\dfrac{x^2}{100}+\dfrac{y^2}{36}=1$$

Solving for the upper half (y ≥ 0):

$$y=\sqrt{36 \left(1-\dfrac{x^2}{100}\right)}$$

To find the height 4 meters from the center, substitute x = 4:

$$y=\sqrt{36 \left(1-\dfrac{4^2}{100}\right)}$$

$$y=\sqrt{30.24}\approx 5.5 \ \text{m}$$

Final Results:

• Arch equation (upper semi-ellipse): \(y=\sqrt{36 \left(1-\dfrac{x^2}{100}\right)}\)
• Height 4 meters from the center: \(y \approx 5.5 \ \text{m}\)