Answer to Question #255564 in Analytic Geometry for Joshua

Question

Direction: Using the information given on each item, write the

standard form equation of each ellipse.

1. Vertices: (12, -4), (-14, -4)

Foci: (11, -4), (-13, -4)

2. Vertices: (6,22), (6, -4)

Foci: (6, 14), (6,4)

3. Foci: (0,9), (-10,9)

Co-vertices: (-5, -21), (-5, -3)

4. Foci: (5,-8), (5,-8)

Co-vertices: (0,4), (0, -20) 5. Center: (6, 2)

Vertex: (6, -4) Co-vertex: (9,2)

Answer 1

1.

The y-coordinates of the vertices and foci are the same, so the major axis is parallel to the x-axis. Thus, the equation of the ellipse will have the form;

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

First, we identify the center. The center is halfway between the vertices, (12, -4), (-14, -4). Applying the midpoint formula, we have:

$\begin{aligned}\left(k,h\right)&=\left(\dfrac{12+\left(-14\right)}{2},\dfrac{-4+\left(-4\right)}{2}\right) \\ &=\left(-1,-4\right) \end{aligned}$

Next, we find a². The length of the major axis, 2a, is bounded by the vertices. We solve for a by finding the distance between the x-coordinates of the vertices.

$\begin{aligned}2a&=-14-\left(12\right)\\ 2a&=-26\\ a&=-13\end{aligned}$

a² = 169

Now we find c². The foci are given by (k±c, h). So, (k−c, h) = (11, −4) and (k+c, h) = (−13, -4). We substitute k = −3 using either of these points to solve for c.

$k+c = -13\\ -1+ c = -13\\ c = -12$

c² = 144

Next, we solve for b² using the equation c² = a² − b².

$c² = a²-b²\\ 144 = 169- b²\\ b² = 25\\ b= 5$

$\dfrac{{\left(x+1\right)}^{2}}{{169}}+\dfrac{{\left(y+4\right)}^{2}}{{25}}=1$

2.

The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form;

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

(h, k) which is the midpoint is;

(h, k) = (6, 9)

Next, we find a². The length of the major axis, 2a, is bounded by the vertices. We solve for a by finding the distance between the y-coordinates of the vertices.

$2a = -4-22 \\ 2a = -26\\ a = -13\\ a² = 169$

$k + c = 4\\ 9+c = 4\\ c = -5\\ c² = 25$

$c² = a²- b²\\ 25 = 169 -b²\\ b² = 144\\ b = 12$

$\dfrac{{\left(x-6\right)}^{2}}{{144}}+\dfrac{{\left(y-9\right)}^{2}}{{169}}=1$

3.

The equation of an ellipse is $\dfrac{(x-h)^2}{a^2} +\dfrac{(y-k)^2}{b^2}=1$ for a horizontally oriented ellipse and $\dfrac{(x-h)^2}{b^2} +\dfrac{(y-k)^2}{a^2} =1$ for a vertically oriented ellipse.

where (h, k) is the center and the distance c from the center to the foci is given by $a^2-b^2=c^2$. a is the distance from the center to the vertices and b is the distance from the center to the co-vertices.

The center of the ellipse is half way between the vertices. Thus, the center (h, k) of the ellipse is (0, 0) and the ellipse is vertically oriented.

a is the distance between the center and the vertices, so a = c is the distance between the center and the foci, so c = 4

$\dfrac{(x-9)^2}{20^2} +\dfrac{(y-12)^2}{5^2} =1$

4.

The equation of an ellipse is$\dfrac{(x-h)^2}{b^2} +\dfrac{(y-k)^2}{a^2} =1$

where (h, k) is the center and the distance c from the center to the foci is given by $a²+b²= c²$. a is the distance from the center to the vertices and b is the distance from the center to the co-vertices.

The center of the ellipse is half way between the vertices. Thus, the center (h, k) of the ellipse is (0, 0) and the ellipse is vertically oriented.

i.e h = 0, k = 0

a which is the x axis point is 4, while b is 20

The equation becomes

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

5.

Center: (6, 2) Vertex: (6, -4) Co-vertex: (9,2)

from the image above,

vertices = (6, -4), (6, 8) => -4+12

covertices = (9, 2) | (3, 2) => 9-3-3

semimajor axis length = 6

semiminor axis length = 6/3 = 2

h = 6

k = 2

$a² -b² = c²\\ 6² -b² = 5²\\ b² = 9\\ b = 3$

$\therefore$ the equation becomes;

$\dfrac{(x - 6)^2 }9+ \dfrac{(y - 2)^2) }{36}= 1$