Answer:
Step-by-step explanation:
Given p + 1 = 2[tex]x^{2}[/tex] → 1 = 2[tex]x^{2}[/tex] - p
[tex]p^{2} + 1 = 2y^{2}[/tex]
[tex]p^{2} + (2x^{2} - p) = 2y^{2}[/tex]
[tex]p^{2} - p + (2x^{2} - 2y^{2} ) = 0[/tex]
The above equation is quadratic in p.
p = [ -(-1) ± [tex]\sqrt{(-1)^{2} - 4* 1*(2x^{2} - 2y^{2} )[/tex] ] ÷ 2
p = [1 ± [tex]\sqrt{1 - 8(x^{2}-y^{2} )}[/tex]] ÷ 2
