Answer
See below
Work Step by Step
Given: $25x^2+49y^2=1225\\\frac{x^2}{49}+\frac{y^2}{25}=1$
The equation is in standard form.
We can see $a=7, b=5$
The denominator of the $x^2-term$ is greater than that of the $y^2-term$, so the major axis is horizontal.
The vertices of the ellipse are at $(\pm a,0)=(\pm 7,0)$. The co-vertices are at $(0,\pm b) = (0,\pm 5)$. Find the foci.
$c^2=a^2-b^2=7^2-5^2=24$
so $c=2\sqrt 6$
The foci are at $(\pm 2\sqrt 6,0)$.
