Help with this problem needed thanks: Find y in terms of a, x, b: x^2/a^2 - y^2/b^2 = 1

I would do a step before what the infamous topsquark suggested. I would first find y2/b2 and then use the above hint.

By the way, no one on this forum will solve your problem for you. After all, this is a math help forum.
Math is not a spectator sport, so you need to be involved in solving your problem.
Please show us your work and where you seem to be getting stuck.
 
This is what I've done so far...

x^(2)/a^(2) - y^(2)/b^(2) =1

x^(2)/a^(2) - 1 = y^(2)/b^(2)

x^(2)b^(2)/a^(2) - b^(2) = y^(2)

Idk what to do now bc it seems a bit weird to just square root everything
 
BanaMath said:
This is what I've done so far...

x^(2)/a^(2) - y^(2)/b^(2) =1

x^(2)/a^(2) - 1 = y^(2)/b^(2)

x^(2)b^(2)/a^(2) - b^(2) = y^(2)

Idk what to do now bc it seems a bit weird to just square root everything
Click to expand...

I think square-rooting both sides (remembering to put a "plus-minus" on the side opposite the [imath]y[/imath]) is probably the way to go. (The graph of the equation is that of an hyperbola, which is not a function but a relation [much like a circle's equation], so you're gonna be stuck with a messy result.)
 
BanaMath said:
This is what I've done so far...

x^(2)/a^(2) - y^(2)/b^(2) =1

x^(2)/a^(2) - 1 = y^(2)/b^(2)

x^(2)b^(2)/a^(2) - b^(2) = y^(2)

Idk what to do now bc it seems a bit weird to just square root everything
Click to expand...
Unless you want to try to put this into some kind of parametric form (I wouldn't recommend it), taking the square root is exactly what you need to do! You are over-thinking it. :)

-Dan
 
BanaMath said:
x^(2)b^(2)/a^(2) - b^(2) = y^(2)

Idk what to do now bc it seems a bit weird to just square root everything
Click to expand...
Can you put into words what seems weird about that? There may be something we can help you see differently.

Your goal is to express y in terms of a, b, and x, and the square root will do just that (as long as you keep in mind that there will be two square roots, positive and negative).
 
Dr.Peterson said:
Can you put into words what seems weird about that? There may be something we can help you see differently.

Your goal is to express y in terms of a, b, and x, and the square root will do just that (as long as you keep in mind that there will be two square roots, positive and negative).
Click to expand...
It just doesn't seem like it is in its simplest form written like this (in the picture below, answer is on the left).
 

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BanaMath said:
It just doesn't seem like it is in its simplest form written like this (in the picture below, answer is on the left).
Click to expand...

I'll agree with you there; the result does not look "simple". But they only asked you to "solve for [imath]y[/imath]", which you've done. You're good to go!
 
BanaMath said:
It just doesn't seem like it is in its simplest form written like this (in the picture below, answer is on the left).
Click to expand...
1693758097041.png
Technically, you haven't solved for y, since y isn't alone on one side; and you omitted a squaring in the denominator. I would put your answer as [math]y=\pm\sqrt{\frac{b^2x^2}{a^2}-b^2}.[/math]
Personally, I would take the square root first, and then multiply by b: [math]y=\pm b\sqrt{\frac{x^2}{a^2}-1},[/math] which feels a little simpler. And someone else might like [math]y=\pm \frac{b}{a}\sqrt{x^2-a^2}.[/math]
But simplification (whatever you choose to consider simple) is not necessary for solving. That's a separate issue, and largely aesthetic. Different forms can be equally correct!
 
BanaMath said:
What if they asked me to solve for y in terms of x, a, b, and 1?
Click to expand...
As Dr Peterson said, you can have 1 since it is not a variable. But you thought that you could. So don't you see that the answer would be the next to last equation in post #11?

Now why can't you have 1 in your question?
What y is a function of is the variables that you would to have to the know the values of to get a value for y. We know what 1 equals and for example what 11.457 equals so you don't need to be given that information to evaluate what y equals in the equation y=xb/11.457 + 1. To find a value for y, you need to know what x and b equals. Hence, y is a function of x and b, NOT a function of x, b, 11.47 and 1.
 
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