Elimination Math Problem: Solving x and y Equations

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The discussion focuses on proving the equation x^2/a^2 + y^2/b^2 = 1 given x = 2at/(1+t^2) and y = b(1-t^2)/(1+t^2). The initial approach involves squaring both equations and attempting to combine them. Participants suggest simplifying each term and manipulating the numerators since they share a common denominator. Ultimately, the left-hand side simplifies correctly to equal 1, confirming the original equation. The solution emphasizes algebraic manipulation and substitution for clarity.
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Homework Statement



If x=2at/(1+t^2) and y=b(1-t^2)/(1+t^2), show that x^2/a^2+y^2/b^2=1



The Attempt at a Solution



I've tried squaring both equations: xt^2=4a^2t^2/(1+2t^2+t^4)

y^2=b^2(1-2t^2t^4)/(1+2t^2+t^4)

Now I've tried adding x^2 and y^2: 4a^2t^2+b^2(1-2t^2+t^4)/(1+2t^2+t^4)

Am I able to cancel any of this down now?
 
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Try solving the y equation for t and substituting that t into the x equation.
 
I think you'll find it easier if you add x2/a2 to y2/b2.

Its made fairly easy since the denominators are the same and thus its just a manipulation of the numerator.
 
First simplify each term x^2/a^2 and y^2/b^2.

Then expand the numerator of y^2/b^2. Add it to x^2/a^2, which has the identical denominator.

Can you get the rest?
 
Last edited:
Oh yes thanks! x^2/a^2+y^2/b^2=4t^2+1-2t^2+t^4/1+2t^2+t^4
So the left hand side cancels to give 1.
Thanks again!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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