solvable for p : y= x(p + √1+p^2)
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it may help you
Step-by-step explanation:
Yes solving with respect to p is possible and we obtain
p=2y±4y2−4ax2−−−−−−−−−√2x=y±y2−ax2−−−−−−−√x
but all also depends form the context of your specific problem, from what p represents, form what you are requested to find.
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