Question

A. Write BOTH the standard and general form of the equation of the specified ellipse. (20 points)

a. Vertices are at (5, -7) and (5, 5) and foci are at (5, -5) and (5, 3)

b. Vertices at (8, -1) and (0, -1) and whose minor axis has length 2.

(Check Examples 1 to 5 - basis for writing standard form)

(Check Examples 8 and 9 - basis for converting standard to general form)

B. Give the coordinates of the center, vertices, covertices, foci, orientation, lengths of major

and minor axis of each equation of ellipse. Then, sketch the graph. (20 points)

a. (x−3)^2/36 + (y+2)^2/100= 1

b. 9x^2 + 25y^2 − 18x + 100y − 116 = 0

Solve, show the complete solution and follow the format below for your final answers:

Orientation: Length of Major axis:

Center: Length of Minor axis:

Foci: Graph:

Vertices:

Co-vertices

Answer 1

$\displaystyle 1.\,\,\, \textsf{The center is located midway between}\\ \textsf{the vertices on the major axis:}\\ \textsf{center}\,\,\left(\frac{5 + 5}{2}, \frac{-7 + 5}{2}\right) = (5, -1) \\ \begin{aligned} \textsf{length of major axis}\,\, &= 12 = 2a \\ \therefore a &= 6 \end{aligned} \\ \begin{aligned} c &= \textsf{length of foci} \\&=\textsf{distance from center to a focus} \\&= \sqrt{4^2} = 4 \end{aligned} \\ c^2 = 16\,\, a^2 = 36 \\ b^2 = a^2 - c^2 = 36 - 16 = 20\\ \textsf{Standard equation of the ellipse:}\\ \frac{(x - 5)^2}{20} + \frac{(y + 1)^2}{36} = 1 \\ \textsf{General form of the ellipse:}\\ 9x^2 + 5y^2 - 90x + 10y + 50 = 0 \\ 2.\,\,\, \textsf{The center is located midway between}\\ \textsf{the vertices on the major axis:}\\ \textsf{center}\,\,\left(\frac{8 + 0}{2}, \frac{-1 - 1}{2}\right) = (4, -1) \\ \begin{aligned} \textsf{length of major axis}\,\, &= 8 = 2a \\ \therefore a &= 4 \end{aligned}\\ \begin{aligned} \textsf{length of minor axis}\,\, &= 2 = 2b \\ \therefore b &= 1 \end{aligned}\\ \textsf{Standard equation of the ellipse:}\\ \frac{(x - 4)^2}{16} + (y + 1)^2 = 1 \\ \textsf{General form of the ellipse:}\\ x^2 - 8x + 16(y + 1)^2 = 0\\ 3(a).\\ \frac{(x - 3)^2}{36} + \frac{(y + 2)^2}{100} = 1\\ a^2 = 36,\,\, a = 6\\ b^2 = 100,\,\, b = 10\\ \textsf{Since}\,\, a < b,\,\, \textsf{the major axis}\\ \textsf{is the}\,\, y-\textsf{axis}\\ \textsf{Length of Major axis}\,\, = 2 \times 10 = 20 \\ \textsf{Length of Minor axis}\,\, = 2 \times 6 = 12 \\ c = \sqrt{100 - 36} = \sqrt{64} = \pm8\\ \begin{aligned} \textsf{Foci}\,\, &= (3, - 2 - 8) \,\, \textsf{or}\,\, (3, - 2 + 8) \\&= (3, -10) \,\, \textsf{or}\,\, (3, 6) \end{aligned} \\ \begin{aligned} \textsf{Vertices}\,\, &= (3, - 2 -10) \,\, \textsf{or}\,\, (3, - 2 + 10) \\&= (3, -12) \,\, \textsf{or}\,\, (3, 8) \end{aligned} \\ \textsf{Center}\,\, = (3, - 2)\\ 3(b).\\ 9x^2 + 25y^2 - 18x + 100y - 116 = 0 \\ 9(x^2 - 2x) + 25(y^2 + 4y) = 116 \\ 9(x^2 - 2x + 1) + 25(y^2 + 4y + 4) = 116 + 100 + 9 = 225\\ 9(x - 1)^2 + 25(y + 2)^2 = 225\\ \frac{(x - 1)^2}{25} + \frac{(y + 2)^2}{9} = 1\\ a^2 = 25,\,\, a = 5\\ b^2 = 9,\,\, b = 3\\ \textsf{Since}\,\, a > b,\,\, \textsf{the major axis}\\ \textsf{is the}\,\, x-\textsf{axis}\\ \textsf{Length of Major axis}\,\, = 2 \times 5 = 10\\ \textsf{Length of Minor axis}\,\, = 2 \times 3 = 6 \\ c = \sqrt{25 - 9} = \sqrt{16} = \pm 4\\ \begin{aligned} \textsf{Foci}\,\, &= (1-4, - 2) \,\, \textsf{or}\,\, (1 + 4, - 2) \\&= (-3, -2) \,\, \textsf{or}\,\, (5, -2) \end{aligned} \\ \begin{aligned} \textsf{Vertices}\,\, &= (1 - 5, - 2) \,\, \textsf{or}\,\, (1 + 5, -2) \\&= (-4, -2) \,\, \textsf{or}\,\, (6, -2) \end{aligned} \\ \textsf{Center}\,\, = (1, - 2)\\$3(a).

3(b).