College Algebra Problems With Answers
Equation of Ellipse

Multiply all terms of the equation by 25 to obtain:

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

The above equation may be written in the standard form \(\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} = 1\) as follows:

\[ \dfrac{x^{2}}{\left(\dfrac{3}{5}\right)^{2}} + \dfrac{y^{2}}{\left(\dfrac{1}{3}\right)^{2}} = 1 \]

With \(a = \dfrac{3}{5}\) and \(b = \dfrac{1}{3}\) (since \(a > b\)). The major axis is the x-axis and its length is equal to \(2a = \dfrac{6}{5} = 1.2\).

• a) Divide all terms of the equation by 32 to obtain: \[ \dfrac{x^{2}}{4} + \dfrac{y^{2}}{16} = 1 \] The above equation may be written as: \[ \dfrac{x^{2}}{b^{2}} + \dfrac{y^{2}}{a^{2}} = 1 \] with \(a = 4\) and \(b = 2\) (since \(a > b\)). Hence the major axis is the y-axis and the minor axis is the x-axis. The length of the major axis is \(2a = 8\) and the length of the minor axis is \(2b = 4\).
• b) The vertices are on the major axis at the points \((0, a) = (0, 4)\) and \((0, -a) = (0, -4)\).
• c) The foci are on the major axis at the points \((0, c)\) and \((0, -c)\) such that: \[ c^{2} = a^{2} - b^{2} = 16 - 4 = 12 \] Hence the foci are at the points \((0, 2\sqrt{3})\) and \((0, -2\sqrt{3})\).

Both the focus and the vertex lie on the y-axis, which means that the major axis is the y-axis. The equation of the ellipse has the form:

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

\(a\) is the distance from the center of the ellipse to a vertex and is equal to 6. \(c\) is the distance from the center of the ellipse to a focus and is equal to 4. Also \(a\), \(b\) and \(c\) are related as follows:

\[ b^{2} = a^{2} - c^{2} = 36 - 16 = 20 \implies b = 2\sqrt{5} \]

The equation of the ellipse is given by:

\[ \dfrac{x^{2}}{20} + \dfrac{y^{2}}{36} = 1 \]

From the coordinates of the foci, \(c = 5\) and the major axis is the y-axis. From the length of the major axis (\(2a = 14\)), we obtain \(a = 7\). Also:

\[ b^{2} = a^{2} - c^{2} = 7^{2} - 5^{2} = 49 - 25 = 24 \]

The equation of the ellipse is given by:

\[ \dfrac{x^{2}}{24} + \dfrac{y^{2}}{49} = 1 \]

The length of the major axis is 10, hence \(a = 5\), and the equation may be written as:

\[ \dfrac{x^{2}}{25} + \dfrac{y^{2}}{b^{2}} = 1 \]

We now use the fact that the point \((3 , \dfrac{16}{5})\) lies on the graph of the ellipse to find \(b^{2}\):

\[ \dfrac{3^{2}}{25} + \left(\dfrac{16}{5}\right)^{2} \cdot \dfrac{1}{b^{2}} = 1 \implies \dfrac{9}{25} + \dfrac{256}{25b^{2}} = 1 \]

Multiply through by \(25b^{2}\):

\[ 9b^{2} + 256 = 25b^{2} \implies 16b^{2} = 256 \implies b^{2} = 16 \implies b = 4 \]

Write the equation as follows:

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

• a) Solve the equation for \(x\) and select the solution for which \(x\) is negative (to the left of the y-axis): \[ 0.2x^{2} = 0.2 - 0.6y^{2} \implies x^{2} = 1 - 3y^{2} \implies x = -\sqrt{1 - 3y^{2}} \]
• b) Solve the equation for \(y\) and select the solution for which \(y\) is negative (below the x-axis): \[ 0.6y^{2} = 0.2 - 0.2x^{2} \implies y^{2} = \dfrac{1}{3} - \dfrac{1}{3}x^{2} \implies y = -\sqrt{\dfrac{1}{3} - \dfrac{1}{3}x^{2}} \]

• a) Center and Axes: Center at \((h, k) = (1, -4)\) with \(a = 4\) and \(b = 3\). The major axis is the vertical line \(x = 1\) and the minor axis is the horizontal line \(y = -4\). Length of major axis = \(2a = 8\), length of minor axis = \(2b = 6\).
• b) Vertices: Vertices on the major axis are at \((1, -4 + 4) = (1, 0)\) and \((1, -4 - 4) = (1, -8)\).
• c) Foci: \(c = \sqrt{a^{2} - b^{2}} = \sqrt{16 - 9} = \sqrt{7}\). Foci are at \((1, -4 + \sqrt{7})\) and \((1, -4 - \sqrt{7})\).

The center of the ellipse is the midpoint of the two foci, located at \(\left(\dfrac{-1 + 3}{2}, 0\right) = (1, 0)\).

\(c\) is the distance from a focus to the center, hence \(c = 2\).

The length of the minor axis is 2 (\(2b = 2\)), hence \(b = 1\).

Calculate \(a^{2}\):

\[ a^{2} = b^{2} + c^{2} = 1^{2} + 2^{2} = 5 \]

Since the foci lie on the x-axis, the major axis is horizontal. The equation of the ellipse is:

\[ \dfrac{(x - 1)^{2}}{5} + \dfrac{y^{2}}{1} = 1 \]

Rewrite the parametric equations:

\[ \dfrac{x}{6} = \sin(t) \quad \text{and} \quad \dfrac{y}{4} = \cos(t) \]

Square both sides of the equations:

\[ \left(\dfrac{x}{6}\right)^{2} = \sin^{2}(t) \quad \text{and} \quad \left(\dfrac{y}{4}\right)^{2} = \cos^{2}(t) \]

Use the trigonometric identity \(\sin^{2}(t) + \cos^{2}(t) = 1\):

\[ \dfrac{x^{2}}{36} + \dfrac{y^{2}}{16} = 1 \]

Here, \(a = 6\) and \(b = 4\). Center is at \((0, 0)\).

• Major axis: x-axis, length = \(2a = 12\)
• Minor axis: y-axis, length = \(2b = 8\)

Group terms and complete the square for \(x\) and \(y\):

\[ (4x^{2} - 16x) + (3y^{2} + 18y) = -31 \] \[ 4(x^{2} - 4x + 4) + 3(y^{2} + 6y + 9) = -31 + 4(4) + 3(9) \] \[ 4(x - 2)^{2} + 3(y + 3)^{2} = -31 + 16 + 27 = 12 \]

Divide by 12 to write in standard form:

\[ \dfrac{(x - 2)^{2}}{3} + \dfrac{(y + 3)^{2}}{4} = 1 \]

• a) Center: \((2, -3)\)
• b) Axes & Lengths: Here \(a^{2} = 4 \implies a = 2\) (vertical major axis), \(b^{2} = 3 \implies b = \sqrt{3}\). Major axis is the line \(x = 2\) with length \(2a = 4\). Minor axis is the line \(y = -3\) with length \(2b = 2\sqrt{3}\).
• c) Vertices: \((2, -3 + 2) = (2, -1)\) and \((2, -3 - 2) = (2, -5)\).
• d) Foci: \(c = \sqrt{a^{2} - b^{2}} = \sqrt{4 - 3} = 1\). Foci are at \((2, -3 + 1) = (2, -2)\) and \((2, -3 - 1) = (2, -4)\).