If I have the following problem:
Vertices: (1,5),(1,-3) and c=3
With the given vertices I know that h=1
I first try to solve for k to finish solving for my center. To this I can use the following format:
(h,k+-a) or (h,k+-b)
If I do not know whether the given vertices are from a major/minor axis, so how do I know whether I should substitute for a or b when solving for k?
In any ellipse, by definition, the minor and major axis have the center of the ellipse as their midpoint. Recalling that the midpoint between $2$ points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$ \left(\frac{x_1 +x_2}{2},\frac{y_1 +y_2}{2}\right) $$ we see that the center of the ellipse has to be at $$ C = \left(\frac{1+1}{2}, \frac{5 -3}{2}\right) = (1,1) $$ So, the equation of our ellipse has the form $$ \frac{(x-1)^2}{a^2} + \frac{(y-1)^2}{b^2} = 1 \tag{1} $$ We'll also note that the distance between the 2 vertices given is $$ d\left[(1,-3),(1,5)\right] = \sqrt{(1-1)^2 + (5-(-3))^2} = 8 $$ Now, to find out if these vertices correspond to the major or minor axis, we recall that, since the vertices need to be on the ellipse, they need to satisfy equation $(1)$ is we plug them in. Choosing one of the vertices, let's say $(1,5)$, by substituting in $(1)$ we get \begin{align} &\frac{(1-1)^2}{a^2} + \frac{(5-1)^2}{b^2} = 1\\ \implies&\frac{4^2}{b^2} = 1 \\ \implies &b = 4 \tag{2} \end{align} And recalling the relation $$ c^2 = a^2 - b^2 $$ we can find the value of $a$ to be \begin{align} a &= \sqrt{c^2 + b^2}\\ &\overset{\color{blue}{(2)}}{=}\sqrt{c^2 + 16}\\ &= \sqrt{9 + 16}\\ & = 5 \end{align} From here we get that the major axis has to have length $2a = 2(5) = 10$ and the minor axis has lenght $2b = 2(4) = 8$. So since the distance between our vertices is $8$, it has to be the case that they correpond to vertices on the minor axis.
Another way of looking at it is as follows. If we just guess that the vertices correspond to the major axis, recalling that the distance between them is $8$ we see that from this assumption we get $a = \frac{8}{2} = 4$. But using the relation $c^2 = a^2 -b^2$ we get that $b$ has length $$ b = \sqrt{a^2 -c^2} = \sqrt{4^2 - 3^2} = \sqrt{7} $$ And since we now know both $a$ and $b$, we see by equation $(1)$ that the equation of this assumed ellipse is given by $$ \frac{(x-1)^2}{4^2} + \frac{(y-1)^2}{\sqrt{7}^2} = 1 $$ However, if we plug in one of the vertices (let's say $(1,5)$) into the equation above we get that \begin{align} &\frac{(1-1)^2}{4^2} + \frac{(5-1)^2}{\sqrt{7}^2} = 1 \\ \implies & \frac{4^2}{7} = 1 \\ \implies & 16 = 7 \end{align} since this is clearly a contradiction, the conclusion is that the point $(1,5)$ is not on the ellipse! But this is a problem, since by it being a vertex it needs to be on the ellipse. Hence the solution where the points correspond to the major axis is impossible.
In conclusion, the condition that $c=3$ forces the vertices to be of the minor axis, and if we were to assume that they correspond to a major axis, we would get an impossible solution. We can visually verify this by plotting the 2 equations we get from assuming each case
where the green ellipse is given by the equation $$ \frac{\left(x-1\right)^{2}}{5^{2}}+\frac{\left(y-1\right)^{2}}{4^{2}}=1 $$ and the red one is given by $$ \frac{\left(x-1\right)^{2}}{4^{2}}+\frac{\left(y-1\right)^{2}}{\sqrt{7}^2}=1 $$ both of them satisfy that $c=3$, that their center is at $(1,1)$ and that one of their axis has length $8$, but only one of the 2 solutions passes through the vertices we're given.