Question #255564

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)


1
Expert's answer
2021-11-12T15:40:48-0500

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;

(xk)2a2+(yh)2b2=1\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:

(k,h)=(12+(14)2,4+(4)2)=(1,4)\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.

2a=14(12)2a=26a=13\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=131+c=13c=12k+c = -13\\ -1+ c = -13\\ c = -12

c² = 144


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

c2=a2b2144=169b2b2=25b=5c² = a²-b²\\ 144 = 169- b²\\ b² = 25\\ b= 5


(x+1)2169+(y+4)225=1\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;

(xh)2b2+(yk)2a2=1\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=4222a=26a=13a2=1692a = -4-22 \\ 2a = -26\\ a = -13\\ a² = 169


k+c=49+c=4c=5c2=25k + c = 4\\ 9+c = 4\\ c = -5\\ c² = 25


c2=a2b225=169b2b2=144b=12c² = a²- b²\\ 25 = 169 -b²\\ b² = 144\\ b = 12


(x6)2144+(y9)2169=1\dfrac{{\left(x-6\right)}^{2}}{{144}}+\dfrac{{\left(y-9\right)}^{2}}{{169}}=1



3.

The equation of an ellipse is (xh)2a2+(yk)2b2=1\dfrac{(x-h)^2}{a^2} +\dfrac{(y-k)^2}{b^2}=1 for a horizontally oriented ellipse and (xh)2b2+(yk)2a2=1\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 a2b2=c2a^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


(x9)2202+(y12)252=1\dfrac{(x-9)^2}{20^2} +\dfrac{(y-12)^2}{5^2} =1


4.

The equation of an ellipse is(xh)2b2+(yk)2a2=1\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=c2a²+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

x242+y2202=1\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

a2b2=c262b2=52b2=9b=3a² -b² = c²\\ 6² -b² = 5²\\ b² = 9\\ b = 3


\therefore the equation becomes;


(x6)29+(y2)2)36=1\dfrac{(x - 6)^2 }9+ \dfrac{(y - 2)^2) }{36}= 1

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