Question

What is the length of the major axis?
(x-7)^2/4 + (y+3)^2/16=1

Answer 1

\(\bf \cfrac{(x-7)^2}{4}+\cfrac{(y+3)^2}{16}=1\implies \cfrac{(x-7)^2}{2^2}+\cfrac{(y+3)^2}{4^2}=1\\ \quad \\
\implies \cfrac{(x-h)^2}{a^2}+\cfrac{(y-k)^2}{b^2}=1\)

Answer 2

hmmm is really a vertical ellipse.... but anyhow, that's how you'd get the major axis length either way