Question

What is the standard form equation of an ellipse that has vertices (18,1) and (−10,1) and co-vertices (4,14) and (4,−12)?

Answer 1

Answer: $\frac{(x-4)^2}{196} +\frac{(y-1)^1}{169} =1$

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

$\frac{(x-h)^2}{a^2} +\frac{(y-k)^2}{b^2} =1$ where (h,k) are the coordinates of the center.

To find the center, find the midpoint of the major axis $(\frac{x1+x2}{2} ,\frac{y1+y2}{2} )$

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:

$\frac{(x-4)^2}{196} +\frac{(y-1)^1}{169} =1$

hope this helps.