I kinda see...
What I meant was:
As x approaches infinity,
Y= (b*infinity)/a * sqroot (1 - a^2/infinity^2) = infinity
However the video I was watching had considered the sqroot as 1 when x approaches infinity, which I understand. But they didn’t substitute it in the outside (bx/a)
Solving for y, you get to y= +/- (b/a)x * square root of(x^2 - a^2)
If x is approaching infinity, then the a^2 is insignificant. Therefore:
y = +/-(b/a) * square root of((x^2)).
This leads to y= (b/a)x.
Remember that x is not infinity itself, it's x approaches infinity. You should be clear from what I typed before, but if you wish, you can view the proof below:
The limit as x approaches infinity of (b/a square-root-of(x^2 - a^2) - bx/a)= 0
Here is a proof of this that I found from a textbook:
lim{x approaches infinity}(b/a (sqrt(x^2 - a^2) - x))
= b/a lim{x approaches infinity}(sqrt(x^2 - a^2) - x)
= b/a lim{x approaches infinity} (sqrt(x^2 - a^2) - x)(sqrt(x^2 - a^2) + x) divided by (sqrt(x^2 - a^2) + x
= b/a lim{x approaches infinity} (x^2 - a^2)^2 - x^2 divided by sqrt(x^2 - a^2) + x
= b/a lim{x approaches infinity} x((1 - a^2/x^2)^2 - 1) divided by x(sqrt(1 - a^2/x^2) + 1
= b/a lim[x approaches infinity] (1 - a^2/x^2)^2 - 1 divided by sqrt(1 - a^2/x^2) + 1
= b/a lim{x approaches infinity}(1)^2 - 1 divided by sqrt(1) + 1
=0/2
=0
Edit: Sorry I just realized that I added too much spacing