Find the equation of an ellipse with one vertex at (3,1) and a minor axis that is parallel to the y axis with a length of 10 and an endpoint located at (16,6)

Question

I have watched many videos and am still confused.

Answer 1

using our standard notation of
(x-h)^2 / a^2 + (y-k)^2 /b^2 = 1

we know 2b = 10 or b = 5
My rough sketch shows that (3,1) must be the bottom of the minor axis
in order for the vertex of the major axis to be at (16,6)
so the centre must be at (3,6) by adding 5 to the vertical distance from (3,1)
the distance from that centre to (16,6) is 13, or we can say a = 13

equation:
(x - 3)^2 / 13^2 + (y - 6)^2 / 25 = 1

(x-3)^2 / 169 + (y - 6)^2 / 25 = 1

Answer 2

i did this as an answer and mt teacher counted it incorrect. the denominaters are right but the numerators are not.