Respuesta :
Answer: [tex]\frac{(x-4)^2}{196} +\frac{(y-1)^1}{169} =1[/tex]
Step-by-step explanation:
You first need to determine if the ellipse is horizontal or vertical, which can be found by the length of the major axis.
The vertices are at (18,1) and (-10,1). The distance is 28 units. (horiz)
The co-vertices are at (4,14) and (4,-12) The distance 26 units (vert)
The major axis is horizontal, so the standard form of the ellipse is
[tex]\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2} =1[/tex] where (h,k) are the coordinates of the center.
To find the center, find the midpoint of the major axis [tex](\frac{x1+x2}{2} ,\frac{y1+y2}{2} )[/tex]
The center is at (4,1)
Since the major axis is 2a, a would be 28 divided by 2=14 squared : 196
The minor axis is 2b, so b would e 26/2=13 squared: 169
Inserting the center and a,b values into the standard form eq:
[tex]\frac{(x-4)^2}{196} +\frac{(y-1)^1}{169} =1[/tex]
hope this helps.
