Question

Y^2+x^2p^2-2xyp=4/p^2

Answer 1

$$p^2x^2-2pxy+y^2=\dfrac{4}{p^2}$$ $$(y-px)^2=\dfrac{4}{p^2}$$

Or

$$y=px\pm\dfrac{2}{p}$$

Each equation is in Clairaut's form. Hence changing $p$ to the arbitrary constant $c,$ the required solution is

$$(y-cx)^2=\dfrac{4}{c^2}, c \text{ is an arbitrary constant}$$