0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.