In order a0->a5: 1/3, 1, 3, 9, 27, 81
All you're doing is plugging the x-value in. So for a2, x=1, so y=3¹ which means y=3.
Remember that
If the given coordinates of the vertices and foci have the form (0,10) and (0,14)
then
the transverse axis is the y-axis
so
the equation is of the form
(y-k)^2/a^2-(x-h)^2/b^2=1
In this problem
center (h,k) is equal to (0,4)
(0,a-k)) is equal to (0,10)
a=10-4=6
(0,c-k) is equal to (0,14)
c=14-4=10
Find out the value of b
b^2=c^2-a^2
b^2=10^2-6^2
b^2=64
therefore
the equation is equal to
<h2>(y-4)^2/36-x^2/64=1</h2><h2>the answer is option A</h2>
