So i have this equation: 36x^2-121y^2=49 which i rearranged to 36x^2/49-121y^2/49=1 so i want to go from here and find the eccentricity, foci, asymptotes
but i'm getting myself confused. since my equation is of the form x^2/a^2-y^2/b^2=1
both a and b would be which makes my eccentricity e=sqrt(1+49/49) but this is at odds with the software i'm using to practice this.
which part am i getting myself tied up with? you!

Well, as the equation form is x^2/a^2-y^2/b^2=1 you have to have in each numerator only x^2 and y^2 but you have 36x^2 and 121y^2, so you have to work more for getting the equation form and i have it for you:
First divide your equation by 36:    36x^2/36-121y^2/36=49/36
you get  x^2-121y^2/36=49/36
Then divide that by 121:   x^2/121-y^2/36=49/4356
 Now divide this by 49/4356:    x^2/(49/36)-y^2/(49/121)=1
Finally sqrt and square for denominators and we will get the equation form:
x^2/(7/6)^2-y^2/(7/11)^2=1
From here you have a=7/6 and b=7/11
Now excentricity is e=(sqrt(a^2+b^2))/a "you have a mistake in this formula"
e=(sqrt((7/6)^2+(7/11)^2)/(7/6) = 1.139 ≈ 1.14 =57/70

Lol idk ask google. ejie