# Finding explicit functions given implicit relationships of x(t),y(t) and t

I'm not sure I used the proper terminology in the title of the question, but here goes. We know that :

$$x^2+y^2=(a+b)t^2+3$$

$$x^2y^2-2=abt^4+(2a+b)t^2$$

where $$a,b>0$$.

How do we prove that :

$$x^2=bt^2+2$$

$$y^2=at^2+1$$

I tried solving for $$x^2$$ in the first equation and then plugging it into the second equation. However, that brings $$y^4$$ and $$y^2$$ terms. I tried completing the square and ended up with this huge square root looking nothing like the solutions above.

Any ideas?

## 1 Answer

To solve such questions easily, it is imperative to simplify them as much as possible, for ease of solving.

We can simplify the above equations by writing them as $$x^2+y^2 = A$$ $$x^2y^2 = B$$ Then, as $$y=\frac{B}{x^2}$$, substituting this in the 1st equation will give us a quadratic in $$x^2$$. We can easily solve this by the quadratic formula, and by substituting the values of $$A$$ and $$B$$, we get the desired result.