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What is an easy method or trick to obtain equation of a parabola from $x$ and $y$ equations? For instance, from $$ x = -2 t_1 t_2(t_1+t_2)$$ and $$ y = {(t_1+t_2)^2-t_1t_2}+(1/2), $$ where $t_1t_2=-1$, how can I obtain the equation of the parabola $$ x^2 = 2(2y-3)$$?

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    $\begingroup$ Let $S=t_1+t_2$ and $P=t_1t_2$. Remember that $(t_1+t_2)^2=S^2-2P$ and solve for $S$ and $P$. Then you can make a substitution. $\endgroup$ – zar May 22 '18 at 7:33
  • $\begingroup$ @zar do you mean t1^2+t2^2 = S^2 - 2P? I can substitute -1 for P, but how can I solve it? $\endgroup$ – Hema May 22 '18 at 7:37
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    $\begingroup$ Yes, $t_1^2+t_2^2 = (t_1+t_2)^2-2t_1t_2=S^2-2P$. So you will have that $x = -2PS$ and $y=S^2-2P-P+1/2$. $\endgroup$ – zar May 22 '18 at 7:44
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    $\begingroup$ I'm sorry, my note is useless: you don't have $t_1^2+t_2^2$ but only $(t_1+t_2)^2$. I have written an answer. $\endgroup$ – zar May 22 '18 at 8:24
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Easy way, straight forward:

1)$x^2=4(t_1+t_2)^2$, or

$x^2/4 = (t_1+t_2)^2$, used $t_1t_2=-1.$

Substitute into your $y$ - equation:

2)$y= x^2/4 +1 +1/2;$

$4(y-3/2)=x^2,$ or

$x^2 = 2(2y-3).$

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Let $P=t_1t_2$ and $S=t_1+t_2$, then $(t_1+t_2)^2=S^2$. By substitution one obtains: $$ \begin{cases} x=-2PS\\ y=S^2-P+\frac{1}{2} \end{cases} $$ This is a general method which works with symmetric equations, even if you don't know the value of $P$.

In you excercise, since $P=-1$, you have $$ \begin{cases} x=2S\\ y=S^2+1+\frac{1}{2}=S^2+\frac{3}{2} \end{cases} $$

You can substitute $S=x/2$ from the first equation into the second one: $$ y=\frac{x^2}{4}+\frac{3}{2} $$ which is equivalent to $$ 4y=x^2+6, $$ so $x^2=4y-6=2(2y-3)$.

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