Question

Find the standard form of the equation of the ellipse with the given characteristics.
vertices: (- 2, 2), (- 2, 16)
minor axis of length 8

Answer 1

ANSWER CHOICES:
(x-2)^2/16+(y+9)^2/49=1
(x+2)^2/16+(y-9)^2/49=1
(x+2)^2/49+(y-9)^2/16=1
(x-9)^2/49+(y+2)^2/16=1
(x+9)^2/49+(y-2)^2/16=1

Answer 2

The length of the major axis is 2a. The length of the minor axis is 2b. The center is at the midpoint of the major axis.

Answer 3

Use that information to find a, b and the center (h,k) and plug into this equation:
\[\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\]

Answer 4

notice the vertices of the ellipse
(-2, 2) and (-2, 16)

notice the x-coordinate doesn't change, but the y-coordinate does
that means the ellipse major axis is the y-axis, or is "vertical"

keep in mind that center of the ellipse, is half-way between the vertices
half-way between (-2, 2) and (-2, 16), let's use the midpoint formula for kicks :)
\(\bf \left(\cfrac{(-2)+(-2)}{2}, \cfrac{(2)+(16)}{2}\right) \implies (-2, 9)\)
so the center is at (-2, 9)
the distance from the center to either vertex is "a"
over the "y" is moved from (-2, 2 ) to (-2, 9), that is 7 units
thus a = 7

the minor axis is given, that is, 8 units, thus b = 8

so
(h, k) = (-2, 9) # center
a = 7 # major axis unit
b = 8 # minor axis uniit

see above as Mertsj showed, the equation for the ellipse, and put your values in :)