Texas HS Precalculus Question of the Day

$\frac{(x-4)^2}{9}+\frac{(y+1)^2}{25}=1$

$\frac{(x-4)^2}{25}+\frac{(y+1)^2}{9}=1$

$\frac{(x+4)^2}{25}+\frac{(y-1)^2}{9}=1$

$\frac{(x-4)^2}{100}+\frac{(y+1)^2}{36}=1$